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

Given 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?

Now the angle between the surface of slope and horizontal is Θ

then work down the slope

 W = m x g x sinΘ (sinΘ = opp / adj)

 W = m x g x 1/49

 W = mg / 49

We know that a = W / m (from Newton's 2nd law of motion)

                       a = m x g / 49 / m

                       a = g / 49

                     a = 9.8 / 49

                     a = 0.2 m / s^2

Now to calculate speed, using Newton's third equation

           v^2 = u^2 + 2 a s

           v^2 = 0 + 2 x 0.2 m/s^2 x 49 m

           v^2 = 0.4 x 49 m^2 /s^2

          v^2 = 19.6 m^2 /s^2

           v = √19.6 m^2/s^2

           v = 4.427 m/s

So acceleration is 0.2 m/s^2 and speed is 4.427 m/s