How to Calculate the Minimum Force Needed to Move a Box on a Ramp with Friction?

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To calculate the minimum force needed to move a 3.00 kg box on a 20-degree ramp with a static friction coefficient of 0.700, one must consider the forces acting on the box. For part (a), when the force is applied perpendicular to the slope, the force must overcome both the gravitational component acting down the ramp and the frictional force, which is influenced by the normal force. In part (b), when applying force horizontally, the calculation becomes more complex as it involves resolving the horizontal force into components that affect both the normal force and the frictional force. Understanding the relationship between these forces is crucial for determining the minimum force required in both scenarios. The discussion emphasizes the importance of correctly applying physics principles to solve for the necessary force.
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A box with a mass of 3.00 kg is at rest on a ramp that is at an angle of 20 degrees with the horizontal. The coefficient of static friction between the box and the ramp is 0.700. Use g = 9.80 m/s2. You now want to make the box move by applying a force. To start the box moving, what is the minimum force you need to apply if your force is directed ...

(a) perpendicular to the slope?

(b) horizontally?
 
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i got part (a) but I'm not sure why it works. mgcos20(mgsin20/static coefficient)

i still don't know how to do part (b) though.
 

Thread 'Mousetrap Car Energy efficiency'

For simple comparison, I think the same thought process can be followed as a block slides down a hill, - for block down hill, simple starting PE of mgh to final max KE 0.5mv^2 - comparing PE1 to max KE2 would result in finding the work friction did through the process. efficiency is just 100*KE2/PE1. If a mousetrap car travels along a flat surface, a starting PE of 0.5 k th^2 can be measured and maximum velocity of the car can also be measured. If energy efficiency is defined by...
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