1-the quanity Fd/t is a- the kinetic energy of the object b- the potential energ
ID: 2300028 • Letter: 1
Question
1-the quanity Fd/t is
a- the kinetic energy of the object
b- the potential energy of the object
c-the work done on the object by the force
e-the power supplied to the object by the force
2-according to the work kinetic energy theorem, if the net work done on an object is negative, then the object's kinetic energy
a- decreases
b- remains the same
c-increase
d-is zero
3-a 500 kg elevator is pulled upward with a constant force of 5500 N for a distance of 50m what is the work done by the 5500 N force?
a)2.75*10^5J
b)-2.45* 10^5J
c)3.00*10^4J
D)-5.20*10^5J
4- an arrow of mass 20 g is shot horizontally into a bale of hay, striking the hay with a velocity of 60 m/s is penetrates a depth of 20 cm before stopping . what is the average stopping force acting on the arrow ?
a)45 N
b)90 N
c)180N
d)360
Calculate the work done by a force F = 2xi + 4y j, around a closed path, first going from P1(1,1) to P2(4,4) along path a and returning along path b. A pendulum of length 50 cm is filled 30 cm away from the vertical axis and release, from rest What. Will be its speed at the bottom of its swing? The kinetic friction force between a 60.0-kg object and a horizontal surface is 50.0 N. If the initial speed object is 25.0 m/s. what distance will it slide before coming to a stop? 15.0 m 30.0 m 375 m 750 m In any and all collisions of short duration and for which it is true that no external forces act on the ct participants kinetic energy is conserved. both momentum and kinetic energy are conserved. neither momentum nor kinetic energy is conserved. the relative velocities before and after impact are equal and oppositely directed, e. momentum is conserved.Explanation / Answer
1) e-the power supplied to the object by the force
2) a- decreases
3) a)2.75*10^5J as W = force * distance = 5500 *50
4) as force * distance = change in KE
>> force = .5 *.020*60^2 / .20 = 180 N
c)180N
6) A) 0 J
7) V= sqrt (2*g*h ) = sqrt (2*9.8 * .1) = 1.4 m/s option (D)
8) x* F = KE >>> x= 375 m option (C)
9) option (e)momentum is conserved
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