Question

A truck starts from rest at the top a slope which is 1 m high and 49 m long. What is the acceleration and it's speed at the bottom of the slope assuming friction is negligible?

Answer 1

`a = 0.2 m/s^2`
`v = 4.4 m/s`

Explanation:

Acceleration:
The truck's weight is m*g. Weight, w, is a force. If the angle between the slope's surface and horizontal is `theta`, then the component of w that points down the slope is

`w_"ds" = m*g*sintheta`

It might help to draw a side view of the situation. Since sine is opposite/adjacent,

`w_"ds" = m*g*(1 m)/(49 m) = (m*g)/49`

From Newton's 2nd Law

`a = w_"ds"/m = ((cancel(m)*g)/49)/cancel(m) = g/49 = (9.8 m/s^2)/49 = 0.2 m/s^2`

Speed:
Use the formula

`v^2 = u^2 + 2*a*d = 0 + 2*0.2 m/s^2*49 m = 19.6 m^2/s^2`

`v = sqrt(19.6 m^2/s^2) = 4.4 m/s`

I hope this helps,
Steve