Question

A 10 kg frictionless cart is pushed at a constant force of 5.0 N for a distance of 10 m. The work done on the cart is 50J.

Answer 1

1) Final kinetic energy of the cart: B) 50 J

2) Final speed of the cart: C) 3.2 m/s

3) Height reached along the ramp: A) 0.5 m

Explanation:

1)

We can solve this part of the problem by using the work-energy theorem, which states that the work done on an object is equal to the kinetic energy gained by the object itself. Mathematically:

$W=K_f - K_i$

where

W is the work done

$K_f$ is the final kinetic energy

$K_i$ is the initial kinetic energy

In this problem, the work done on the cart is

W = 50 J

And assuming it starts from rest, its initial kinetic energy is zero:

$K_i = 0$

Therefore, the final kinetic energy is:

$K_f = K_i + W=0+50=50 J$

2)

The kinetic energy of an object is the energy possessed by an object due to its motion; it is calculated as

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

For the cart in this problem, we have:

K = 50 J is its final kinetic energy

m = 10 kg is the cart

Therefore, solving the formula for v, we find its speed:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(50)}{10}}=3.2 m/s$

3)

We can think this problem in terms of conservation of energy. In fact, as the cart rolls up the ramp, its kinetic energy is converted into gravitational potential energy, which is given by

$PE=mgh$

where

m is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

h is the heigth of the cart

When the cart reaches the maximum height, all the kinetic energy has been converted into potential energy, so we can write:

$K=PE\\\frac{1}{2}mv^2=mgh$

Re-arranging,

$h=\frac{v^2}{2g}$

And since we know the initial speed of the cart along the ramp,

v = 3.2 m/s

we can find the maximum height reached along the ramp:

$h=\frac{3.2^2}{2(9.8)}=0.5 m$

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