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Find the work done on a box on an inclined plane

  • Thread starter popo902
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  • #1
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Homework Statement



a constant force Fa of magnitude 85.0 N is applied to a 4.0 kg shoe box at angle φ = 53.0°, causing the box to move up a frictionless ramp at constant speed. How much work is done on the box by Fa when the box has moved through vertical distance h = 0.45 m?

heres the picture to explain where the phi comes from http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c07/pict_7_32.gif

Homework Equations


W=F dot D
or W = FDcostheta


The Attempt at a Solution


i know that if an object is going at constant speed it's net force is zero
i tried solving for D which gave me .45/sintheta
then i got y comp of the weight force = 4gsinthet
the x comp of the wieght cut the phi angle in two = 53 - theta
and using Fa and angle (53-thet) i got the force that balance the 4g sin = 85sin(53 - thet)

then i used 4g sinthet = 85sin(53 - thet) to find theta because the net force is zero
i then put the theta i got into the D equation above.
I then found the x and y components of D and the x and y compnents of F using the angle( 53-thet) and did the dot product of that which came out to be 59.51455753 J
and it's wrong....
the answer is 17.658 J...supposedly
can someone tell me what i'm doing wrong? am i using the wrong angle or am i just going abot the problem wrong from the beginning? help please
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
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5
How much work is done on the box by Fa when the box has moved through vertical distance h = 0.45 m?
When the work is done on a body, its kinetic energy increases. But in the problem it is stated that the box moves with uniform velocity. How is that?
Next the work done on the box = rise in potential energy of the box = m*g*h
If you substitute the values you get the answer.
 
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