A block of mass 12 kg starts from rest and slides a distance of 7 m down an inclined plane making an angle of 40◦ with the horizontal. The coefficient of sliding friction between block and plane is 0.4. The acceleration of gravity is 9.8 m/s2 . What is the net force on the block along the incline? Answer in units of N.
029 (part 2 of 3) 10.0 points What is the speed of the block after sliding 7 m? Answer in units of m/s.
030 (part 3 of 3) 10.0 points What would be its speed if friction were negligible? Answer in units of m/s.
M*g = 12 * 9.8 = 117.6 N. = Wt. of block.
Fp = 117.6*sin40 = 75.6 N. = Force parallel to incline.
Fn = 117.6*Cos40 = 90.1 N. = Normal force.
Fk = u*Fn = o.4 * 90.1 = 36.0 N = Force of kinetic friction.
A. Fnet = Fp-Fk = 75.6 - 36.0 = 39.6 N. = Net force.
B. Fnet = M*a.
39.6 = 12*a, a = 3.3 m/s^2.
V^2 = Vo^2 + 2a*d. = 0 + 2*3.3*7 = 46.2
V = 6.8 m/s.
C. Fnet = Fp-Fk = 75.6-0 = 75.6 N.
Fnet = M*a.
75.6 = 12a, a = 6.3 m/s^2.
V^2 = Vo^2 + 2a*d = 0 + 2*6.3*7 = 88.2,
V = 9.4 m/s.
Well, well, well, sliding down an inclined plane, huh? Let's calculate some forces and speeds!
First, let's determine the net force acting on the block along the incline. We can break down the forces into two components: the gravitational force pulling the block downhill and the friction force trying to slow it down.
The gravitational force can be calculated using: F_gravity = m * g * sin(θ)
where m is the mass, g is the acceleration due to gravity, and θ is the angle of the incline.
Now, the friction force can be determined using: F_friction = m * g * cos(θ) * μ,
where μ is the coefficient of sliding friction.
To find the net force, we subtract the frictional force from the gravitational force: F_net = F_gravity - F_friction.
For the first part of the question, we will calculate the net force.
Calculating each component:
F_gravity = 12 kg * 9.8 m/s^2 * sin(40°) ≈ 76.61 N
F_friction = 12 kg * 9.8 m/s^2 * cos(40°) * 0.4 ≈ 31.42 N
Now, let's find the net force:
F_net = F_gravity - F_friction ≈ 45.19 N
So, the net force on the block along the incline is approximately 45.19 N.
Now, onto the second part of the question: the speed of the block after sliding 7 m.
We can use the equation: v^2 = u^2 + 2as, where u is the initial velocity (0 m/s, as the block starts from rest), a is the acceleration, and s is the distance.
The acceleration can be calculated using the equation: a = F_net / m.
Plugging in the values:
a = 45.19 N / 12 kg ≈ 3.77 m/s^2
v^2 = 0^2 + 2 * 3.77 m/s^2 * 7 m ≈ 52.78 m^2/s^2
Taking the square root of both sides, we find:
v ≈ √52.78 ≈ 7.27 m/s
So, the speed of the block after sliding 7 m is approximately 7.27 m/s.
Finally, onto the third part: What would be the speed if friction were negligible?
If we neglect friction, then the only force acting on the block would be the gravitational force.
Using the same v^2 = u^2 + 2as equation:
v^2 = 0^2 + 2 * 9.8 m/s^2 * 7 m ≈ 137.2 m^2/s^2
Taking the square root of both sides, we find:
v ≈ √137.2 ≈ 11.72 m/s
So, if friction were negligible, the speed of the block after sliding 7 m would be approximately 11.72 m/s.
There you have it, my friend. Forces, speeds, and a whole lot of physics! Who said science can't be entertaining?
To find the net force on the block along the incline, we need to consider the forces acting on it. There are three main forces at play: the gravitational force (mg), the normal force (N), and the force of friction (f).
First, let's calculate the gravitational force acting on the block. The gravitational force is given by the formula F_gravity = mg, where m is the mass of the block (12 kg) and g is the acceleration due to gravity (9.8 m/s²). Therefore, F_gravity = 12 kg × 9.8 m/s² = 117.6 N.
Next, we need to determine the normal force (N). The normal force is the component of the force perpendicular to the inclined plane, and its value can be calculated using N = mgcosθ, where θ is the angle of inclination (40°). Therefore, N = 12 kg × 9.8 m/s² × cos(40°) ≈ 92.4 N.
Now, let's calculate the force of friction (f). The force of friction is given by the formula f = μN, where μ is the coefficient of sliding friction (0.4) and N is the normal force. Therefore, f = 0.4 × 92.4 N = 36.96 N.
To find the net force on the block along the incline, we need to take into account the direction of each force. The gravitational force acts downwards (-117.6 N), the normal force acts perpendicular to the plane (92.4 N), and the force of friction acts opposite to the direction of motion (-36.96 N).
Net force on the block along the incline = F_gravity + f = -117.6 N + (-36.96 N) = -154.56 N.
Therefore, the net force on the block along the incline is approximately -154.56 N.
To determine the speed of the block after sliding 7 m, we can use the equation of motion: v² = u² + 2as, where v is the final velocity, u is the initial velocity (0 m/s as it starts from rest), a is the acceleration, and s is the distance.
The acceleration of the block can be calculated using the net force (F_net) divided by the mass (m) of the block. Therefore, a = F_net / m = -154.56 N / 12 kg ≈ -12.88 m/s².
Now, substituting the obtained values into the equation v² = u² + 2as, we have v² = 0 m/s + 2 × (-12.88 m/s²) × 7 m.
v² = -180.32 m²/s².
Taking the square root of both sides, we find v ≈ -13.43 m/s (negative sign indicates the direction).
Therefore, the speed of the block after sliding 7 m is approximately 13.43 m/s in the opposite direction to the motion.
Lastly, to determine the speed of the block if friction were negligible, we can use the same equation v² = u² + 2as.
Since the force of friction provides the acceleration, if friction is negligible, there would be no acceleration. Thus, a = 0 m/s².
Substituting the values into the equation, v² = 0 m/s + 2 × 0 m/s² × 7 m.
v² = 0 m²/s².
Therefore, the speed of the block if friction were negligible is 0 m/s.