A mass of 6 kg is placed on a frictionless incline which is inclined at 47 degre

Question

A mass of 6 kg is placed on a frictionless incline which is inclined at 47 degrees. A force of 21 Newtons is applied to it directly up the incline and another force is applied to the left (see sketch). Due to these forces, the amount of the normal force on the mass is 133 Newtons. The mass is at rest at the bottom of the incline and after the forces have been applied for 2.8 seconds, the mass has moved all the way up the incline. What is the length (not height) of the incline in meters? The mass never comes off the incline.

Explanation / Answer

Inclination of incline=47o

F1 = 21 N

Total normal force on mass = 133 Newtons

Therefore , F2 sin47 + 6*9.8*cos47=133

F2 = 127 N

Component of F2 along incline = 127cos47=86.63 N(up the incline)

Total force on mass along incline = 21+86.63-(6*9.8*sin47)=64.63 N

acceleration of block = 64.63/6=10.77 m/s2

Length of incline = (0.5*10.77*2.82)=42.2 m

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