Moving a brick

Classical Mechanics Level 3

You are trying to move a $33 \text{ kg}$ brick on the floor by $10$ meters at a constant speed. You push the brick with a constant magnitude of force, using both arms which form an angle of $32^{\circ}$ with the floor. What is the magnitude of the work (in J) you put on the brick if the coefficient of kinetic friction between the floor and brick is $0.2?$

621 \text{ J}

588 \text{ J}

829 \text{ J}

739 \text{ J}

4 solutions

Sudipan Mallick
May 23, 2014

Draw the free body diagram. equating the force equations along y axis- N=mg+Fsin32 frictional force=0.2(Fsin32+mg) now equate force wquation along x axis- Fcos32=frictional force so, on solving the equations we get F=87.146 N now work done=Fcos32 x distance moved =739J

Sagar Kulkarni
Nov 28, 2014

if F is the force offered by the hand, Fcos32 is the magnitude of force in the direction opposite to the friction. and frictional force is uN=0.2* 33 * 9.81. If you equate these two terms, you can get F=73.9 N. W= force * displacement. so, W= 73.9*10= 739J---Ans

Bryan Dellariarte
May 20, 2014

sorry self explanatory and research

A diagram is needed since the problem did not specify push or pull. The problem did not also specify if the angle formed was above or below the horizontal. 588 J would be the solution if you assumed that you pull the brick.

Jay Tio - 7 years ago

but one can try to pull them too if considering that case you will get 588J

Ananya Chattree - 7 years ago

Shriya Mandarapu
May 16, 2014

for there to be constant speed net force in the horizontal direction must be zero..which would give us Fcos32=0.2(Fsin32+mg) values of sin 32 nd cos 32 would be 0.529 and 0.848 respectively solving would give F=87.146N work done=F.S=FScos32=739J

same solution...

Bryan Dellariarte - 7 years ago

Moving a brick

4 solutions

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