Box sliding in an inclined plane?

Question

The 50kg crate is projected down a 30 degree incline with respect to the horizontal with initial speed of 8m/s. Determine the time required for the crate to come to rest and the corresponding x traveled.

The coefficient of kinetic friction is 0.4

I kept getting incorrect answers.

The correct answer is t=6.39 s x=25.6m

Answer 1

Use Newton's 2nd Law of Motion to determine the acceleration of the crate.

F = ma

where

F =  net force acting on the crate

m = mass of the crate = 50 kg

a = acceleration

From the free body diagram of the crate (I trust that you can make a free body diagram),

f - mg(sin 30) = ma

where

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

f = frictional force

By definition,

f = coeff of friction * mg(cos 30) = 0.4mg(cos 30)

Substituting appropriate values,

0.4mg(cos 30) - mg(sin 30) = ma

Since "m" appears on both sides of the equation, then it will simply cancel out. Hence the above equation simplifies to

0.4g(cos 30) - g(sin 30) = a

a = 9.8(0.4*cos 30 - sin 30)

a = - 1.505 m/sec^2

NOTE the negative sign of the acceleration. This simply denotes that the crate is actually slowing down as it is sliding down the incline (due to the force of friction).

The working formula to use is

Vf - Vo = aT

where

Vf = final velocity = 0 (when the crate stops)

Vo = initial velocity = 8 m/sec (given)

a = - 1.505 m/sec^2 (as calculated above)

T = time for the crate to stop along the incline

Substituting appropriate values,

0 - 8 = (-1.505)T

T = 8/1.505 = 5.32 sec.

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Formula to use is

Vf^2 - Vo^2 = 2ax

where

x = distance along the incline crate will travel before coming to a stop

and all the other terms have already been defined previously.

Substituting values,

0 - 8^2 = 2(-1.505)x

x = 64/(2*1.505)

x = 21.26 meters

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My answers are different from what you presented.  

Answer 2

Is the crate pushed from bottom of incline in upward direction? Or, is it dropped from top of incline and allowed to move downward?

Answer 3

Start with a freebody diagram and sum the forces moving normal to the plane

50*9.81*0.4*cos(30 deg)-50*9.81*sin(30  deg)=50*Accel.(or dv/dt)

Integrate both sides to get:

(50*9.81*0.4*.866-50*9.81*0.5)*t= 50*(- 8)

one equation, one unknown