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3 years ago

Calculate the potential energy of a 1200 kg boulder on a cliff 45 m above the ground.

Physics

1 answer:

Softa [21]3 years ago

4 0

Explanation:

Given mass of the object is 1200kg and it's placed at a height of 45m above the ground. As we know that potential energy is ,

$\longrightarrow PE = mgh$

where,

m is the mass of the body .
g is acceleration due to gravity.
h is the height above the ground .

Substituting the respective values ,

$\longrightarrow P.E. = 1200kg * 10m/s^2* 45m$

Multiply ,

$\longrightarrow P.E. = 540000J$

Hence the potential energy is 540000J .

I hope this helps.

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Answer:

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Explanation:

A 3.00-kg crate slides down a ramp. the ramp is 1.00 m in length and inclined at an angle of 30° as shown in the figure. The crate starts from rest at the top, experiences a constant friction force of magnitude 5.00 N, and continues to move a short distance on the horizontal floor after it leaves the ramp. Use energy methods to determine the speed of the crate at the bottom of the ramp.

Solution:

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$W_f=F_f\Delta S\\\\Where\ F_f\ is\ the \ frictional\ force=-5N(the\ negative \ sign\ because\ it\\acts\ opposite\ to \ direction\ of\ motion),\Delta S=slope\ length=1\ m\\\\W_f=F_f\Delta S=-5\ N*1\ m=-5J$

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$W_g=F_g*s*cos(\theta)\\\\F_g=force\ due\ to\ gravity=mass*acceleration\ due\ to\ gravity=3\ kg*9.81\\m/s^2, s=1\ m, \theta=angle\ between\ force\ and\ displacement=90-30=60^o\\\\W_g=3\ kg*9.81\ m/s^2*1\ m*cos(60)=14.72\ J\\\\The\ Kinetic\ energy(KE)=W_f+W_g=14.72\ J-5\ J=9.72\ J\\\\Also, KE=\frac{1}{2} mv^2\\\\9.72=\frac{1}{2} (3)v^2\\\\v=\sqrt{\frac{2*9.72}{3} } =2.55\ m/s$

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