A 20.5kg crate sits at bottom of a 9.5m long ramp that makes an angle 25.6degree
ID: 1346304 • Letter: A
Question
A 20.5kg crate sits at bottom of a 9.5m long ramp that makes an angle 25.6degree with the horizontal. A rope is secured to the top end of the crate and is stretched over a pulley attached to top of ramp. A person on the free end of the rope stands behind the ramp pulling on the rope to move the crate up the ramp. Friction is present between the crate and the ramp. The coefficient of static friction is 0.22 and for kinetic friction is 0.47. A. Move crate at constant speed up ramp. a) The force needed to begin moving the crate up the ramp is ____ N. b) The force needed to move the crate at constant speed up the ramp is ____ N. c) The work done by the person to move the crate from the bottom of the ramp to the top is ____ J. d) The net work done to move the crate to the top of the ramp is _____ J. B. Accelerate box up the ramp e) The force needed to accelerate the crate at 3 m/s2 is ____ N. f) The speed of the crate when it reaches the top of the ramp is ____ m/s
Explanation / Answer
perpendicular to the ramp on crate,
N -mgcos25.6 = 0
N = mgcos25.6
and max. Static friction force = 0.47mgcos25.6
kinetic friction force = 0.22mgcos25.6
along the ramp crate is pulled with constant velcocity so acc. a = 0
a) crate is not moving so static friction will act
F - 0.47mgcos25.6 -mgsin25.6 = 0
F - 0.47 x 20.5 x 9.81 x cos25.6 - 20.5x9.81xsin25.6 = 0
F= 172.13 N
b) once carte started moving now kinetic friction will act.
F - 0.22mgcos25.6 -mgsin25.6 = 0
F - 0.22 x 20.5 x 9.81 x cos25.6 - 20.5x9.81xsin25.6 = 0
F= 126.79 N
c)work done by person = 126.79 x 9.5 =1204.55 J
d) Net work done on crate = change in KE of crate
Net work done = 0 - 0 = 0
e)
Fnet = ma
F - 0.22mgcos25.6 -mgsin25.6 = 20.5 x 3
F = 188.29 N
f) v^2 - u^2 =2ad
v^2 - 0 = 2 x 3 x 9.5
v = 7.55 m/s
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