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Stells [14]
3 years ago
9

A boy pushes a 50 kg wheelbarrow with a velocity of 0.5 m/s. If he uses a force of 10 N, how far does he push

Physics
1 answer:
labwork [276]3 years ago
4 0

Answer:

d=1.25 m

Explanation:

fd=.5mv2-.5mv1; 10(d)=.5(50)(.5)-0; 10(d)=12.5; d=12.5/10; d=1.25 m

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Explanation:

So, The bob is held at angle theta from initial position, P. Find the displacement from P.  

displacement= L sin  Θ. I hope you get it here. The pendulum forms a triangle with length L, the initial position and horizontal displacement.

So,the point is at displacement point the kinetic energy of bob is 0 and when it is released it will have maximum kinetic energy at P. It will hit the box and all he kinetic energy of bob will get transferred into driving force(F.D) of box. Now kinetic energy = Force * displacement.

K.E= 0.5 m V^2= F.D * Lsin Θ.

Find F.D

F.D = (0.5 m V^2) /  Lsin Θ.

Now for the box, F.D - Friction = m(box) a.

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Find the accelaration of box from an equation of motion. ( u=0, find displacement of box,s,time taken and so on)

U get the friction. Now,

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So here u go.. U get the coeff of friction.. I hope am right here.. and made no mistake.. Anyway try it with the values to confirm! ;)

7 0
3 years ago
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