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.

Answer 2

The velocity of the cart for the given constant force is 2.83 m/s.

The given parameters;

• constant force, F = 10 N
• mass of the cart, m = 20 kg
• distance traveled by the cart, d = 8 m

The acceleration of the cart is calculated as follows;

$F = ma\\\\a = \frac{F}{m} \\\\a = \frac{10}{20} \\\\a = 0.5 \ m/s^2$

The velocity of the cart is calculated as follows;

$v^2 = u^2 + 2ad\\\\v^2 = 2ad\\\\v = \sqrt{2ad} \\\\v = \sqrt{2(0.5)(8)}\\\\v = 2.83 \ m/s$

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