Question

A constant 10.0-N horizontal force is applied to a 20.0-kg cart at rest on a level floor. If friction is negligible, what is the speed of the cart when it has been pushed 8.0 m? Assume cart starts from rest.

Answer 1

Answer:

2.83 m/s

Explanation:

Given:

Mass of the cart (m) = 20.0 kg

Initial velocity of the cart (u) = 0 m/s

Final velocity (v) = ? m/s

Displacement of the cart (S) = 8.0 m

Horizontal force acting on the cart (F) = 10.0 N

Surface is frictionless. So, only horizontal force is the force acting on the cart.

Now, as per work-energy theorem, the work done by the net force acting on a body is equal to the change in the kinetic energy of the body.

Here, the work done by the horizontal force is given as:

$W_{net}=FS=(10\ N)(8.0\ m)=80\ Nm$

The change in kinetic energy is given as:

$\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}(20.0\ kg)(v^2-0)\\\\\Delta K=10v^2$

Now, from work-energy theorem:

$\Delta K=W_{net}\\\\10v^2=80\\\\v^2=\frac{80}{10}\\\\v=\sqrt{8}=2.83\ m/s$

Therefore, the speed of the cart when it has been pushed 8.0 m is 2.83 m/s.