Question

A 7.25 kg7.25 kg block is sent up a ramp inclined at an angle ????=28.5°θ=28.5° from the horizontal. It is given an initial velocity ????0=15.0 m/sv0=15.0 m/s up the ramp. Between the block and the ramp, the coefficient of kinetic friction is ????k=0.326μk=0.326 and the coefficient of static friction is ????s=0.593.μs=0.593. How far up the ramp in the direction along the ramp does the block go before it comes to a stop

Answer 1

15.03 m

Step-by-step explanation:

Given:

mass of the block, m = 7.25 kg

Angle, Θ = 28.5°

Initial speed of the block, v₀ = 15 m/s

let the distance traveled by the block be 's'

Now, applying the work energy theorem,

we have

`(m* g*\sin(\theta)* s) + \mu_k* mg* s* cos(\theta) = (1)/(2)* m* v^2`

on substituting the values in the above equation, we get

`(7.25* 9.8*\sin(28.5^o)* s) + 0.326* 7.25*9.8* s* cos(28.5^o) = (1)/(2)* 7.25* 15^2`

or

`33.902* s) +20.35* s = 815.625`

or

`54.252* s = 815.625`

s = 15.03 m

Hence, the block will travel 15.03 m up the ramp