1) A 10.0 kg box is released from rest on a 31.0 degree incline. If the box acce

Question

1) A 10.0 kg box is released from rest on a 31.0 degree incline. If the box accelerates down the slope at a rate of 1.75 m/s2, the coefficient of kinetic friction between the box and the incline must be ______

2) A box slides down an incline with a coefficient of kinetic friction between the box and surface of 0.150. If the box starts from rest, it will take __________ seconds to reach a speed of 20.0 m/s.

3)If a hill has a radius of curvature of 80.0 meters, a car traveling at ________ m/s will 'just' leave the road at the top of the hill.

A 10.0 kg box is released from rest on a 31.0 degree incline. If the box accelerates down the slope at a rate of 1.75 m/s2, the coefficient of kinetic friction between the box and the incline must be A box slides down an incline with a coefficient of kinetic friction between the box and surface of 0.150. If the box starts from rest, it will take seconds to reach a speed of 20.0 m/s. If a hill has a radius of curvature of 80.0 meters, a car traveling at m/s will 'just' leave the road at the top of the hill.

Explanation / Answer

1)

mass m= 10 kg , angle = 31 degrees , acceleration a = 1.75 m/s^2

from newton law normal n = mgcos31 = 10*9.8*cos31 = 84 N

apply newtons law along incline

mgsin31 - un = ma

10*9.8*sin31 - u *84 = 10*1.75

u = (50.47 -17.5) = 0.392

2)

80*9.8 sin35 -0.15*80*9.8*cos35 = 80 a

a = (449.68 - 96.33) /80 = 4.416 m/s^2

given intial velocity u = 0

final v =20

from v = u +at

20 = 4.146 t

time t = 4.52 sec

3)

speed v = sqrt ( 2rg ) =sqrt(2*80*9.8) = 39.59 m/s

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