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

Answer:

15.03 m

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

$(m\times g\times\sin(\theta)\times s) + \mu_k\times mg\times s\times cos(\theta) = \frac{1}{2}\times m\times v^2$

on substituting the values in the above equation, we get

$(7.25\times 9.8\times\sin(28.5^o)\times s) + 0.326\times 7.25\times9.8\times s\times cos(28.5^o) = \frac{1}{2}\times 7.25\times 15^2$

or

$33.902\times s) +20.35\times s = 815.625$

or

$54.252\times s = 815.625$

s = 15.03 m

Hence, the block will travel 15.03 m up the ramp