Two people wish to push a big beer freezer weighing 2000 N up a ramp inclined at

Question

Two people wish to push a big beer freezer weighing 2000 N up a ramp inclined at an angle of 37.0o with the horizontal. The coefficient of kinetic friction between the freezer and the ramp is 0.500. (a) What minimum force must the people exert to slide the freezer up the ramp? (b) What acceleration will the freezer have if it is released and slides down the ramp? (c) If it slides 4.00 m down the ramp and strikes a heavy object, coming to rest in 0.500 m, what average force does it exert on the heavy object?

Explanation / Answer

a)
when sliding up friction will be down the ramp
friction = .5mgcos()

mgsin() acts downwards

minimum force needed = mgsin() + .5mgcos()

mg = 2000 N

= 37

Fmin = 2069.655 N

b)

this case friction acts upwards

acceleration = gsin(37) -.5gcos(37) = 1.98 m/s^2

c)

velocity at the after 4 m slide = sqrt(2*1.98*4) = 3.98

it came to rest after .5 m

deceleration = 3.98^2/2*.5 = 15.84

average force = 2000*15.84/9.8 = 3232.65 N

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