Question

A truck of mass 5 ton is travelling on a horizontal road with 36km/hr stops on travelling 1km after its engine fails suddenly . What fraction of its weight is the frictional force exerted by the road

Answer 1

Answer:

0.5x10^-4

Explanation:

to calculate the value of gravity here is the final formula

μ= 1/(2  x g x d)

g= gravity constant

d= distance

here are the steps to calculate this formula

final velocity - initial velocity = 2 x a x d             equation 1

a can be calculated form

F = ma = -μ x mx g           ( force applied = force stopping)

m will cancel out

a= -μ x g

put this in equation 1

36km/h ( 36 * 1000/3600) to convert it in to m/s

( 1m/s ) inital speed

0- 1m/s = 2 ( -μ x g ) x d

-1m/s = 2 ( -μ) x 10 x 1000 (m)

mins sign will cancel out

and equation will be like

μ= 1/(2 x 10 x 1000)

so the answer will be

1/20000 or 0.00005

or 0.5 x 10 ^ -4

Answer 2

Answer:

Explanation:

I have attached the answer below

For clarity