Question

An object with a mass m slides down a rough 37° inclined plane where the coefficient of kinetic friction is 0.20. If the plane is 10 m long and the mass starts from rest, what will be its speed at the bottom of the plane?

Answer 1

Answer: 9.312 m/s

Explanation:

The friction force (opposite to the motion) is Fa = μ*m*g*cos(α) with μ = kinetic friction. The force that makes the motion is

F = m*g*sin(α).

The Newton's law gives:

F - Fa = m*a

m*g*sin(α) - μ*m*g*cos(α) = m*a

g*sin(α) - μ*g*cos(α) = a so a = 4.335 m/s²

It's a uniformly accelerated motion:

Space

S = 0.5*a*t²

10 = 0.5*a*t²

=> t = 2.148 s

Velocity

V = a*t = 9.312 m/s.

Answer 2

We have that the speed at the bottom of the plane is

$v-9.3m/s$

From the question we are told that:

Angle of slide $\theta =3.7 \textdegree$

Coefficient of kinetic friction $\mu= 0.20$

Length $L=10m$

Generally, the equation for acceleration along the slide is mathematically given by

$a=gsin \theta-\mu cos\theta$

$a=(9.8sin37-0.20*9.8*cos37$

$a=4.33m/s^2$

Therefore

Velocity v is  is mathematically given by

$v=\sqrt{2as}$

$v=\sqrt{2*4.33*10}$

$v-9.3m/s$

In conclusion

The speed at the bottom of the plane is

$v-9.3m/s$

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