Question

A box sits at the top of an incline that is 50.0 cm tall and that makes an angle of 25° with the ground. The box has a mass of 2.0 kg. It is given an initial speed of 3.0 m/s down the incline. If the coefficient of kinetic friction between the box and the incline is 0.10, how long does it take the box to reach the bottom of the incline?

Answer 1

As the box sits at the top of an incline that is 50.0 cm tall and makes an angle of 25° with the ground, The time it takes to reach the bottom of the incline is mathematically given as

t=0.148s

What is the time it takes the box to reach the bottom of the incline?

Generally, the equation for the acceleration  is mathematically given as

$a=g(sin\theta-uscos\theta)$

Therefore

a=(9.8sin-(0.15cos35))

a=4.417

Thereofore

v^2=v0^2+2al

$v^2=V0^2+2ahsin\theta\\\\V^2=(2.0+2(4.417))*60*0.01sin35$

v=2.6533

In conclusion, The time t is

t=v-v0/a

2.65-20/4.417

t=0.148s

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