2. Net force, force components, and friction. A box (mass=35kg) is set on an inc

Question

2. Net force, force components, and friction.

A box (mass=35kg) is set on an incline (angle=35.5o). The coefficient of kinetic friction between the box and the incline is 0.15. The box is let go and starts sliding down. What will be its acceleration? Use the following steps to find your answer: a. Draw a free-body diagram identifying the forces, and their directions. b. Show your calculations of the force components for relevant forces. c. Derive an expression for the normal force in terms of other parameters. d. Derive an expression for the frictional force in terms of other parameters. e. Derive the expression for net force, and solve for the acceleration. Plug in the numbers, with units to arrive at the final answer.

I solved these problems, however I don't know if I am right. Thanks!!!

Explanation / Answer

1.

friction force on the box=0.15*35*9.8*cos(35.5)=41.886 N

component of its weight along the incline=35*9.8*sin(35.5)=199.18 N

then net force=199.18-41.886=157.294 N

acceleration=force/mass=4.494 m/s^2

Q2. forces acting are:

weight of the box downwards

friction force upwards

(you can resolve the weight along the incline and perpendicular to incline)

Q3. normal force=component of weight of the box along perpendicular to the incline

=35*9.8*cos(35.5)=279.24 N

Q4. frictional force=friction coefficient*normal force

=0.15*279.24=41.886 N

Q5. net force=force along incline-friction force

=35*9.8*sin(35.5)-41.886=157.294 N

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