a crate of mass 50kg slides down a 30degree incline. the crate\'s acceleration i

Question

a crate of mass 50kg slides down a 30degree incline. the crate's acceleration is 2.0m/s and the incline is 10m long.

what is the kinetic energy of the crate as it reaches the bottom of the incline ?

how much work is spent in overcoming friction?

what is the magnitude of the frictional force that acts on the crate as it slides down the incline?

what ia the coefficient of kinetic friction between the crate and the incline.

at the base of thr incline there is a horizontal surface with the same coeeficient of kinetic friction how far will the crate slide before coming to rest

Explanation / Answer

a) the kinetic enegy is

K = ( m v2 ) / 2

where, the speed of the crate when reaches the incline is

v2 =2 a1 d

then

K = ( m 2 a1 d ) / 2 = m a1 d

K = 50 Kg * 2.0 m/s2 * 10 m

K = 1000 J

b) consider that

E = W

E - E0 = W

K - U = W

K - m g h = W

K - m g d sin = W

W = 1000 J - 50 Kg * 9.81m/s2 * 10 m * sin( 30° )

W = -1452.5 J

c) the magnitude of the frictional force is

W = - fk d --->   fk = - W / d

fk = - ( -1452.5 J ) / ( 10 m )

fk = 145.3 N

d) the coefficient of kinetic friction is

fk = k N

where

N - m g cos = 0 --->   N = m g cos

then

fk = k m g cos --->   k = fk / ( m g cos )

k = ( 145.3 N ) / ( 50 Kg * 9.81 m/s2 * cos(30°) )

k = 0.34

e) on the horizontal surface

- fk = m a2

- k N = m a2

- k m g = m a2   ---> a2 = - k g

then

Vf2 = v02 + 2 a2 x

0 = v02 - 2 k g x ---> x = v02 / ( 2 k g ) = ( 2 a1 d ) / ( 2 k g )

x = ( 2 * 2.0 m/s2 * 10 m  ) / ( 2 * 0.34 * 9.81 m/s2 )

x = 5.99 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.