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stiv31 [10]
3 years ago
6

Suppose a 48-N sled is resting on packed snow. The coefficient of kinetic friction is 0.10. If a person weighing 660 N sits on t

he sled, what force is needed to pull the sled across the snow at constant speed?
Physics
1 answer:
Annette [7]3 years ago
8 0

Assume the snow is uniform, and horizontal.

Given:

coefficient of kinetic friction = 0.10 = muK

weight of sled = 48 N

weight of rider = 660 N

normal force on of sled with rider = 48+660 N = 708 N = N

Force required to maintain a uniform speed

= coefficient of kinetic friction * normal force

= muK * N

= 0.10 * 708 N

=70.8 N


Note: it takes more than 70.8 N to start the sled in motion, because static friction is in general greater than kinetic friction.


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Answer:

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Explanation:

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work done = ΔK.E ( change in Kinetic energy )  ---- ( 1 )

<em>where :</em>

work done = p * t

                  = 15 * 10^6 watts * ( 1 year ) = 473040000 * 10^6 J

( note : convert 1 year to seconds )

and ΔK.E = 1/2 mVf^2   given ; m = 1200 kg  and initial V = 0

<u>back to equation 1 </u>

473040000 * 10^6  = 1/2 mv^2

Vf^2 = 2(473040000 * 10^6 ) / 1200

∴ Vf = 887918.92 m/s

<u>i) Determine how fast the rocket is ( acceleration of the rocket )</u>

a = Vf / t

  = 887918.92 / ( 1 year )

  = 0.0282 m/s^2

<u>ii) determine distance travelled by rocket </u>

Vf^2 - Vi^2 = 2as

Vi = 0

hence ; Vf^2 = 2as

s ( distance ) =  Vf^2 / ( 2a )

                     = ( 887918.92 )^2 / ( 2 * 0.0282 )

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Why solar panels are usually painted black​
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What is the impulse of a 3 kg object that starts from rest and moves to 20 m/s?
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Answer:

The impulse on the object is 60Ns.

Explanation:

Impulse is defined as the product of the force applied on an object and the time at which it acts. It is also the change in the momentum of a body.

F = m a

F = m(\frac{v_{2}  - v_{1} }{t})

⇒     Ft = m(v_{2} - v_{1})

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From the question, m = 3kg, v_{1} = 0m/s and v_{2} = 20m/s.

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