Refer to the diagram shown below.
μ = the coefficient of dynamic friction between the crate and the ramp.
1. The applied force of F acts over a distance, d.
The work done is F*d.
2. The component of the weight of the crate acting down the ramp is
mg sin(30) = 0.5mg.
The work done by this force is 0.5mgd.
3. The normal force is N = mgcos(30) = 0.866mg.
This force is perpendicular to the ramp, therefore the work done is zero.
4. The frictional force is μN = μmgcos(30) = 0.866μmg.
The work done by the frictional force is 0.866μmgd.
5. The total force acting on the crate up the ramp is
F - mgsin(30) - μmgcos(30) = F - mg(0.5 - 0.866μ)
6. The work done on the crate by the total force is
d*(F - 0.5mg - 0.866μmg)
Hookes law states F=kX where F is the force applied, k is the spring constant, and X is the extension of the spring from its resting point.
Substituting the values in, we get:
F=100*(0.65-0.5)=100*0.15=15N
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Answer:
0.119 s
Explanation:
Given that

Lets take final speed of the softball after covering 4.8 m = V
We know that

Also We know that

Putting the value of V ,U and a in the previous equation We get

Therefore slide time will be 0.119 s
I hope this helps you! The bowling ball would go from potential energy to kinetic energy.