A constant force of 20 N is exerted on a 5 kg block, initially at rest, for a di

Question

A constant force of 20 N is exerted on a 5 kg block, initially at rest, for a distance of 10 m along a horizontal frictionless surface. The block travels for an additional 10 m without the force acting upon it. and then travels up a frictionless hill whose height is 5 m. (See figure below.) How much work does the force do on the block? 400 J 200 J 100 J 50 J How fast is the block moving when the force stops? 80 m/s 8.94 m/s 7.94 m/s 6.94 m/s How fast is the block moving just before it starts up the hill? 80 m/s 8.94 m/s 7.94 m/s 6.94 m/s To what vertical height up the hill does the block travel? 2.08 m 3.08 m 4.08 m It travels over the top (max height would have been 5.08 if the hill had been high enough).

Explanation / Answer

Given that

force F=20 N

mass m=5 kg

distance d=10 m

now we find the work done by force does block moves

the work donw W=F*d=20*10=200 N

now we find the velocity block moving force to stop

F=mv^2/2d

20=5*v^2/2*10

v^2=80

velocity V=8.94

now we find the velocity before the move top of hill

w=1/2mv^2

200=1/2*5*v^2

v^2=80

velocity v=8.94 m/s

now we find the height does block travel

the height h=200/5*9.8=4.08 m

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