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.

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

The given parameters;

The acceleration of the cart is calculated as follows;

The velocity of the cart is calculated as follows;

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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:

The change in kinetic energy is given as:

Now, from work-energy theorem:

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

he made 11 calls in one day

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form