What was the magnitude of the force of friction during the time the block slowed?

Question

A block of mass 2 kg is sliding down an inclined that has a coefficient of kinetic friction of 0.4. The block slows from 2 m/s initially to rest. The incline is at 30 degrees. What was the magnitude of the force of friction during the time the block slowed?

A) 3.4 kg m/s^2

B) 4.0 kg m/s^2

C) 5.1 kg m/s^2

D) 6.8 kg m/s^2

E) 8.3 kg m/s^2

A) 3.4 kg m/s^2 
B) 4.0 kg m/s^2 
C) 5.1 kg m/s^2 
D) 6.8 kg m/s^2 
E) 8.3 kg m/s^2

oldschool · Accepted Answer

Answer is D) Friction force = m*g*0.4*cos30 = 6.8

mukavetz · Answer

The friction force is mu times the component of mg perpendicular to the surface. Find the perpendicular component using geometry. Note that the friction force is max at theta of zero so cos of the angle gives the component you need.

let me know thru comments if you have any trouble.

oubaas · Answer

friction force = m*g*μ*cos 30°
gravitational force = m*g*sin 30°
Energy balance :
1/2*m*Vo^2+m*g*sin 30°*Δx = m*g*μ*cos 30°*Δx
m cross
Vo^2+2*g*sin 30°*Δx = 2*g*μ*cos 30°*Δx
Vo^2 = 2*g*Δx*(μ*cos 30°-sin 30°) 
4 = Δx*19.6*(0.346-0.5)
given data are inconsistent , since (0.346-0.5) is negative and it can't be !!!!....friction coefficient must be higher ....if we use μ = 0.695 instead of 0.4 we get :
4 = Δx*19.6*(0.866*0.695-0.5) = 2.00Δx
Δx = 4.00/2.00 = 2.00 m
stopping time t = 2*Δx/Vo = 2*2/2 = 2.00 sec
acceleration a = (0-V)/t = -2/2 = -1.00 m/sec^2
friction force Ff = m*g*μ*cos 30° = 19.6*0.866*0.695 = 11.80 N

Anonymous · Answer

So many insightful answers here

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What was the magnitude of the force of friction during the time the block slowed?

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