A small block with a mass of m = 0.120 kg is attached to a cord passing through a hole in a frictionless, horizontal surface. The block is originally revolving at a distance of 0.40 m from the hole with a speed of 0.70 m/s.
The cord is pulled from below, shortening the radius of the circle in which the block revolves to 0.10 m. At this new distance, the speed of the block is observed to be 2.80 m/s
a) What is the tension in the cord in the original situation when the block has a speed of: v = 0.70 m/s?
b) What is the tension in the cord in the final situation when the block has speed of:
v = 2.80 m/s?
c) How much work was done by the person who pulled the cord?
a) (m*v^2)/d=(0.12*0.7^2)/0.4=0.147 N
b)(m*v^2)/d=(0.12*2.80^2)/0.1=9.408 N
c) W=deltaKE=1/2mv^2 -1/2mu^2 =1/2 x 0.12 x [(2.8)^2 - (0.70)^2] =0.44 J
did u get the answer?????
a) Ah, the tension in the cord when the block has a speed of 0.70 m/s. Let me calculate that for you. But first, a joke to lighten the mood: Why was the math book sad? Because it had too many problems!
To calculate the tension, we need to understand the centripetal force acting on the block. This force is provided by the tension in the cord. So, using the formula F = m*(v^2)/r, where F is the centripetal force, m is the mass, v is the velocity, and r is the radius, we can calculate the tension:
T = m*(v^2)/r = 0.120 kg * (0.70 m/s)^2 / 0.40 m
Now let's do the math while I summon my calculator... *poof*... Oh no, my calculator flew away; it must have wanted to go on vacation. But don't worry, I can still do the math in my head. Just give me a second... *hums circus music*
Alright, the tension in the cord is approximately 0.147 N. Ta-da!
b) Now for the tension in the final situation when the block has a speed of 2.80 m/s. This time, we're dealing with a shorter radius, but let's calculate it just like before, using the formula T = m*(v^2)/r:
T = 0.120 kg * (2.80 m/s)^2 / 0.10 m
And... let me put on my clown thinking cap... *humming*... Ah, yes, the tension in the cord is approximately 9.792 N.
c) Lastly, how much work was done by the person who pulled the cord? Well, work is equal to the change in kinetic energy. So let's find the change in kinetic energy by using the formula ΔKE = (1/2)*m*(v_f^2 - v_i^2), where v_f is the final velocity and v_i is the initial velocity:
ΔKE = (1/2) * 0.120 kg * (2.80 m/s)^2 - (0.120 kg * (0.70 m/s)^2)
Calculating... carry the one... a little juggling... and the change in kinetic energy is approximately 0.470 J.
Congratulations to the person who pulled the cord! They did approximately 0.470 J of work. I hope that lightened the day for both of you!
To solve this problem, you need to use the concept of centripetal force and work done.
a) To find the tension in the cord in the original situation when the block has a speed of v = 0.70 m/s, we can use the centripetal force equation:
F = (m * v^2) / r
where F is the centripetal force, m is the mass of the block, v is the speed, and r is the radius.
Given:
m = 0.120 kg
v = 0.70 m/s
r = 0.40 m
Plugging in the values into the equation:
F = (0.120 kg * (0.70 m/s)^2) / 0.40 m
F = 0.147 N
Therefore, the tension in the cord in the original situation is 0.147 N.
b) To find the tension in the cord in the final situation when the block has a speed of v = 2.80 m/s, we can use the same centripetal force equation.
Given:
m = 0.120 kg
v = 2.80 m/s
r = 0.10 m
Plugging in the values into the equation:
F = (0.120 kg * (2.80 m/s)^2) / 0.10 m
F = 9.792 N
Therefore, the tension in the cord in the final situation is 9.792 N.
c) To find the work done by the person who pulled the cord, we can use the work-energy theorem:
Work = Change in kinetic energy
The change in kinetic energy can be calculated as:
ΔK = (1/2) * m * (v^2 - u^2)
where ΔK is the change in kinetic energy, m is the mass, v is the final velocity, and u is the initial velocity.
Given:
m = 0.120 kg
v = 2.80 m/s
u = 0.70 m/s
Plugging in the values into the equation:
ΔK = (1/2) * 0.120 kg * ((2.80 m/s)^2 - (0.70 m/s)^2)
ΔK = 0.294 J
Therefore, the work done by the person who pulled the cord is 0.294 J.
To find the answers to the given questions, we can start by using the principles of circular motion and conservation of angular momentum.
a) To find the tension in the cord in the original situation, we can use the concept of centripetal force. The tension in the cord provides the centripetal force that keeps the block in circular motion. The formula for centripetal force is:
F = m * (v^2 / r)
where F is the force, m is the mass, v is the velocity, and r is the radius.
In this case, the block's mass is given as 0.120 kg, the velocity is 0.70 m/s, and the radius is 0.40m. Substituting these values into the formula, we get:
F = 0.120 kg * (0.70 m/s)^2 / 0.40 m
F = 0.147 N
So, the tension in the cord in the original situation when the block has a speed of 0.70 m/s is 0.147 N.
b) To find the tension in the cord in the final situation, we can use the conservation of angular momentum. The angular momentum of the revolving block is constant unless acted upon by an external torque. Mathematically, it can be expressed as:
L = m * v * r
where L is the angular momentum, m is the mass, v is the velocity, and r is the radius.
In this case, the block's mass is given as 0.120 kg, the velocity is 2.80 m/s, and the new radius is 0.10 m. Substituting these values into the formula, we get:
L = 0.120 kg * 2.80 m/s * 0.10 m
L = 0.0336 kg.m^2/s
Since angular momentum is conserved, we can equate the initial and final angular momenta:
m * v_initial * r_initial = m * v_final * r_final
0.120 kg * 0.70 m/s * 0.40 m = 0.120 kg * v_final * 0.10 m
Solving for v_final, we get:
v_final = (0.120 kg * 0.70 m/s * 0.40 m) / (0.120 kg * 0.10 m)
v_final = 2.80 m/s
So, the tension in the cord in the final situation when the block has a speed of 2.80 m/s is also 0.147 N.
c) To find the work done by the person who pulled the cord, we can use the work-energy principle. The work done is equal to the change in kinetic energy of the block.
The initial kinetic energy of the block is given by:
KE_initial = 0.5 * m * v_initial^2
The final kinetic energy of the block is given by:
KE_final = 0.5 * m * v_final^2
The change in kinetic energy is therefore:
ΔKE = KE_final - KE_initial
ΔKE = (0.5 * m * v_final^2) - (0.5 * m * v_initial^2)
Substituting the given values for mass, initial velocity, and final velocity, we get:
ΔKE = (0.5 * 0.120 kg * (2.80 m/s)^2) - (0.5 * 0.120 kg * (0.70 m/s)^2)
ΔKE = 0.756 J
Therefore, the work done by the person who pulled the cord is 0.756 J.