Question

A cart of mass M = 2.40 kg can roll without friction on a level track. A light string draped over a light, frictionless pulley connects the cart to a hanger with mass m = 0.50 kg. The cart is released from rest, allowing it to roll, pulled by the falling hanger.
What is the formula for the acceleration a of the hanger in terms of the tension T, the hanger's mass m, and the acceleration of gravity g? The key for the formula entries assumes the following: T and g are positive numbers, and in the coordinate system up and right are positive, while down and left are negative. Hint: Get a piece of paper and draw careful free body diagrams of the cart and the hanger. Make sure that the vector for the tension force in each diagram has the same length.
Formula for ahanger =
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What is the formula for the acceleration a of the cart in terms of the tension T, and its mass M?
Formula for acart =
Use the two equations you found above to eliminate T to give the acceleration of the cart in terms of M, m and g. (Be careful about the sign of each acceleration!)
Revised formula for acart =
Calculate the the numerical value for the acceleration (Assume g = 9.81 m/s2).
Value of acart =

Answer 1

Answer:

Part a)

$a_{hanger} = g - \frac{T}{m}$

Part b)

$a_{cart} = \frac{T}{M}$

Part c)

$a = \frac{mg}{M + m}$

Part d)

$a = 1.35 m/s^2$

Explanation:

Part a)

For hanger we know that it will have tension force upwards while it has downwards its weight so we will have

$mg - T = ma$

so we have

$a_{hanger} = g - \frac{T}{m}$

Part b)

now for car that is rolling on the floor the net force is given as

$F = Ma$

$T = Ma$

$a_{cart} = \frac{T}{M}$

Part c)

now we know that the cart and the hanger both are connected to each other

so they must have same acceleration

so we will have

$T = Ma$

$mg - Ma = ma$

$a = \frac{mg}{M + m}$

Part d)

now we know that

M = 2.40 kg

m = 0.50 kg

so we will have

$a = \frac{0.50(9.81)}{2.40 + 0.50}$

$a = 1.35 m/s^2$