A cart of mass M1 = 6 kg is attached to a block of mass M2 = 3 kg by a string that passes over a frictionless pulley. The system is initially at rest and the table is frictionless. After the block has fallen a distance h = 1 m:
What is the work Ws done on the cart by the string?
(the cart is on the table attached to a pulley from which a block is hanging)
After the block has fallen 1 m, potential energy of M2 g H = 3*9.8*1 = 29.4 J is converted to kinetic energy of both the cart and the block. Since they will both be travelling at the same speed, and the cart has 2/3 of the total mass, only 2/3 of the potential energy loss acts to accelerate the cart. That would be 19.6 J.
You could get the same result by solving two free body equations to get the tension in the string, and multiplying it by the distance moved.
T = M1 g = M2 a
M2 g -T = M1 a
M2 g = (M1 + M2) a
a = g [M2/(M1 + M2)]
T = M2 a = M2 g * [M2/(M1 + M2)]
= 2/3 * M2 g
d
Well, the work done on the cart by the string can be calculated using the equation:
Work (W) = Force (F) * Distance (d)
In this case, the force acting on the cart is the tension in the string, which we previously found to be 2/3 of the weight of M2. And the distance moved by the cart is the same as the distance the block has fallen, which is 1 m.
So, the work done on the cart by the string is:
W = (2/3 * M2 * g) * d
Plugging in the values, we get:
W = (2/3 * 3 * 9.8) * 1
W = 19.6 J
So, the work done on the cart by the string is 19.6 Joules.
The work done on the cart by the string can be calculated using the formula:
Work (Ws) = Force (F) * Distance (d)
In this case, the force exerted by the string is the tension (T) in the string, and the distance moved is the height (h) the block has fallen.
From the previous calculation, we determined that the tension in the string is 2/3 times the weight of the block:
T = 2/3 * M2 * g
So, the work done on the cart by the string is:
Ws = T * d
Substituting the given values, we get:
Ws = (2/3 * M2 * g) * h
Replacing the values with the given masses and acceleration due to gravity:
Ws = (2/3 * 3 kg * 9.8 m/s^2) * 1 m
Simplifying:
Ws = 19.6 J
The work done on the cart by the string is 19.6 Joules.
To calculate the work done on the cart by the string, we can use the formula:
Work (W) = Force (F) × Distance (d) × cos(θ),
where F is the force applied by the string, d is the distance the cart moves, and θ is the angle between the force and the direction of motion.
In this case, the force applied by the string is equal to the tension (T) in the string. The cart moves a distance of 1 m vertically downward. Since the force and displacement are in the same direction, the angle θ is 0 degrees, and therefore, cos(θ) = 1.
Now, let's calculate the tension in the string:
T = (2/3) × M2 × g,
where M2 is the mass of the block and g is the acceleration due to gravity.
Plugging in the given values:
T = (2/3) × 3 kg × 9.8 m/s²,
T = 19.6 N.
Finally, we can calculate the work done by the string:
W = T × d × cos(θ),
W = 19.6 N × 1 m × 1,
W = 19.6 J.
Therefore, the work done on the cart by the string is 19.6 Joules.