A block of mass m = 2.00 kg rests on the left edge of a block of mass M = 8.00 kg. The

coefficient of kinetic friction between the two blocks is 0.300, and the surface on which the 8.00-
kg block rests is frictionless. A constant horizontal force of magnitude F = 10.0 N is applied to
the 2.00-kg block, setting it in motion across the top of the lower block. If the distance across the
larger block is 3.00 m (from front edge of smaller block to rightmost edge of larger block),
(a) how long will it take the smaller block make it to the right side of the 8.00-kg block. (b) How
far will the 8.00-kg block move in this time?

m1=2 kg, m2=8 kg, μ=0.3, F=10 N.


For m1:
m1•g=N,
m1•a1=F-F(fr) = F- μ•N=F- μ•m1•g.
a1=F/m1 - μ•g = 10/2 -0.3•9.8 = 2.06 m/s²

For m2:
m2•a2=F(fr)
a2=F(fr)/m2= μ•m1•g/m2 = 0.3•2•9.8/8 = 0.735 m/s².

Distances
x1=a1•t²/2,
x2=a2•t²/2,

x1=x2+L,
a1•t²/2 = a2•t²/2 + L,
Solve for t
t=2.13 s.
x2=a2•t²/2= 1.67 m

given, m1 = 2kg ; m2 = 8kg; coefficient of friction = 0.3; F = 10 N

For m1:
m1*g = N
m1*a1 = F-F(friction) = F- (coefficient of friction)*N = F- μ*m1*g
a1 = F/m1 - μ*g = 10/2 - 0.3*9.8 = 2.06 m/s²

For m2:
m2*a2 = F(friction)
a2 = F(friction)/m2= μ*m1*g/m2 = 0.3*2*9.8/8 = 0.735 m/s²
Distance
x1 = a1*t^2/2
x2 = a2*t^2/2
x1 = x2+L,
a1*t^2/2 = a2*t^2/2 + L
Solving for t,
t = 2.13 s
x2 = a2*t^2/2 = 1.67 m
Answer

Well, well, well, it seems like we have some physics homework here! Don't worry, I'm here to clown around and help you out.

(a) To find out how long it will take the smaller block to make it to the right side of the 8.00-kg block, we need to consider the forces involved. The only horizontal force acting on the smaller block is the applied force of magnitude F = 10.0 N, and there's also the force of friction between the two blocks.

The frictional force can be found using the equation F_friction = coefficient_friction * normal_force.
The normal force can be calculated as mass * gravity, where gravity is approximately 9.8 m/s^2.

Once we have the frictional force, we can use Newton's second law (F = ma) to find the acceleration of the smaller block. Since the forces are acting in the horizontal direction, we can just focus on the x-axis.

Once we have the acceleration, we can use the equation of motion: x = x0 + v0t + 1/2at^2 to solve for time, where x is the distance (which is 3.00 m), x0 is the initial position (0), v0 is the initial velocity (also 0), a is the acceleration, and t is the time we're looking for.

(b) Now, to find out how far the 8.00-kg block moves in this time, we can use the same equation of motion, but with the acceleration being 0 (since there is no horizontal force acting on it).

I hope this helps, and remember, physics doesn't have to be a clown fiesta. Just take it step by step and you'll be acing those problems in no time!

To solve this problem, we can use Newton's second law and the equations of motion to find the time it takes for the smaller block to reach the right side of the larger block, as well as the distance the larger block moves.

Let's calculate the acceleration of the smaller block first:

1. Calculate the frictional force acting on the smaller block:
Frictional force (f) = coefficient of kinetic friction (μ) * Normal force
Normal force = mass (m) * acceleration due to gravity (g)
Normal force = 2.00 kg * 9.8 m/s^2 = 19.6 N
f = 0.300 * 19.6 N = 5.88 N

2. Calculate the net force acting on the smaller block:
Net force (Fnet) = Applied force (F) - Frictional force (f)
Fnet = 10.0 N - 5.88 N = 4.12 N

3. Calculate the acceleration of the smaller block:
Fnet = m * a
4.12 N = 2.00 kg * a
a = 2.06 m/s^2

Now, let's calculate the time it takes for the smaller block to reach the right side of the larger block:

4. Calculate the distance that the smaller block travels across the larger block:
Distance (d) = 3.00 m

5. Use the equation of motion:
d = (1/2) * a * t^2
3.00 m = (1/2) * 2.06 m/s^2 * t^2

6. Solve for time (t):
t^2 = (2 * 3.00 m) / 2.06 m/s^2
t^2 = 2.91 s^2
t ≈ √2.91 s ≈ 1.70 s

Therefore, it will take approximately 1.70 seconds for the smaller block to make it to the right side of the 8.00-kg block.

Now, let's calculate the distance the 8.00-kg block moves in this time:

7. Use the equation of motion:
d = v * t
Distance (d) = Unknown velocity (v) * 1.70 s

8. We need to find the velocity of the 8.00-kg block.
To do this, we need to consider the forces acting on the 8.00-kg block.

- The only external horizontal force acting on the 8.00-kg block is the frictional force
between the two blocks. As the surface on which the 8.00-kg block rests is frictionless,
there is no external force acting on the 8.00-kg block.

- According to Newton's first law, an object at rest or moving with constant velocity has
a net force of 0 Newtons. Hence, the acceleration of the 8.00-kg block is 0 m/s^2.

- Since the acceleration is 0, the velocity of the 8.00-kg block remains constant.

9. Therefore, the distance the 8.00-kg block moves is given by:
Distance (d) = Velocity (v) * 1.70 s
d = 0 * 1.70 m
d = 0 m

Thus, the 8.00-kg block does not move in this time period.

To solve this problem, we can break it down into several steps:

Step 1: Calculate the force of friction between the two blocks.

The force of friction can be calculated using the equation:

\( f_{\text{friction}} = \mu \cdot f_{\text{normal}} \)

where:
\( \mu \) is the coefficient of kinetic friction between the two blocks, which is given as 0.300
\( f_{\text{normal}} \) is the normal force between the two blocks

We know that the normal force is equal to the weight of the upper block, so:

\( f_{\text{normal}} = m_{\text{upper}} \cdot g \)

where:
\( m_{\text{upper}} \) is the mass of the upper block, 2.00 kg
\( g \) is the acceleration due to gravity, approximately 9.8 m/s²

Substituting the values into the equation, we get:

\( f_{\text{normal}} = 2.00 \, \text{kg} \cdot 9.8 \, \text{m/s²} \)

Now we can calculate the force of friction:

\( f_{\text{friction}} = 0.300 \cdot (2.00 \, \text{kg} \cdot 9.8 \, \text{m/s²}) \)

Step 2: Calculate the acceleration of the system.

The net force acting on the upper block can be calculated using the equation:

\( f_{\text{net}} = f_{\text{applied}} - f_{\text{friction}} \)

where:
\( f_{\text{applied}} \) is the applied force on the upper block, 10.0 N

Now we can calculate the acceleration using Newton's second law:

\( f_{\text{net}} = m_{\text{upper}} \cdot a \)

Solving for acceleration:

\( a = \frac{{f_{\text{net}}}}{{m_{\text{upper}}}} \)

Substituting the values, we get:

\( a = \frac{{10.0 \, \text{N} - f_{\text{friction}}}}{{2.00 \, \text{kg}}} \)

Step 3: Calculate the time taken by the smaller block to reach the right side of the larger block.

To calculate the time taken, we can use the equation of motion:

\( x = x_0 + v_0t + \frac{1}{2}at^2 \)

where:
\( x \) is the distance traveled by the upper block, 3.00 m
\( x_0 \) is the initial position, 0 m
\( v_0 \) is the initial velocity, 0 m/s
\( a \) is the acceleration calculated in step 2
\( t \) is the time taken

Rearranging the equation, we get:

\( t^2 + \frac{2x_0}{a}t + \frac{2x}{a} = 0 \)

Solving this quadratic equation using the quadratic formula, we can find the value of \( t \).

Step 4: Calculate the distance moved by the larger block.

Since the upper block is moving across the top of the lower block, the distance moved by the larger block will be the same as the distance traveled by the smaller block. Therefore, the distance moved by the larger block will also be 3.00 m.

So, in summary:

(a) Calculate the force of friction between the two blocks using the equation \( f_{\text{friction}} = \mu \cdot f_{\text{normal}} \).
(b) Calculate the acceleration of the system using the equation \( f_{\text{net}} = f_{\text{applied}} - f_{\text{friction}} \) and \( a = \frac{{f_{\text{net}}}}{{m_{\text{upper}}}} \).
(c) Calculate the time taken by the smaller block to reach the right side of the larger block using the equation of motion.
(d) Calculate the distance moved by the larger block, which will be the same as the distance traveled by the smaller block.