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A 100N object is on an incline at 370. The coefficients of static and kinetic friction are both 0.30.

a.) What is the smallest force parallel to the incline needed to keep the object from sliding down?
b.) What parallel force is required to keep  the object moving up the incline at a constant speed?
c.) Assume the force, F, up the incline is 94N. What is the acceleration of the object?
d.) From part c, how far will it move in 10s?
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2 Answers

a) Apply Newton's second law in the direction paralle to the incline
F+f-mgsin37 = 0
=> F = mgsin37 - 0.3*mgcos37 = 100(sin37-0.3cos37) = 36. N
 
b) Apply Newton's second law in the direction paralle to the incline
F-f-mgsin37 = 0
=> F = mgsin37 + 0.3*mgcos37 = 100(sin37+0.3cos37) = 84.N
 
c) 94-84 = ma = mg(a/g)
a = 10g/100 = 0.98 m/sec^2
 
d) d = (1/2)at^2 = (1/2)(0.98)(10^2) = 49 m
Hey Sierra -- the visuals on this stem from the common 3-4-5 /37 right triangle ...
the 100N weight pushes 60N downramp and 80N normal to the ramp ...
friction opposes motion at 30% of the 80N or 24N ...  
 
a) to hold in place, we need 36N plus the 24N friction to counter the downramp 60N ...
b) constant speed upramp needs to overcome the 60N down and 24N friction = 84N ...
c) at 94N, we have extra 10N upramp moving a 10kg mass at F/m of 1 m/s/s ...
d) after 10s, speed is 10m/s ... ave speed of 5m/s for 10s moves ~ 50m ... Regards :)