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IRINA_888 [86]
4 years ago
8

A bicyclist increases her speed from 4.0 m/s to 6.0 m/s. The combined mass of the bicyclist and the bicycle is 55 kg. How much w

ork did the bicyclist do in increasing her speed?
A. 11 J
B. 28 J
C. 55 J
D. 550 J
Physics
2 answers:
11Alexandr11 [23.1K]4 years ago
4 0
Kinetic energy of the horse and rider = 1/2 (mass) (speed)²

At 4.0 m/s :  KE = 1/2 (55kg) (4 m/s)²  =  440 joules

At 6.0 m/s :  KE = 1/2 (55kg) (6 m/s)²  =  990 joules

She increased the kinetic energy of herself and her vehicle by 550 joules,
so she must have put at least that much work into it.
almond37 [142]4 years ago
3 0
The answer is D) 550j
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A 2000-kg truck is being used to lift a 400-kg boulder B that is on a 50-kg pallet A. Knowing the acceleration of the rear-wheel
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Explanation:

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when the truck moves 1 m to the left, the boulder is B and pallet A are raised 0.5 m, then,

a_{A} =  0.5 m/s^{2} (upward) , a_{B} =  0.5 m/s^{2} (upward)

Let T be tension in the cable

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                                  = 2T- (400 + 50)*(9.81 m/s^{2}) = (400 + 50)*(0.5 m/s^{2})

                        T = 2320N

Truck:  M_{R} = ∑(M_{R})eff: = -N_{f} (3.4m) + m_{T} (2.0m) - T (0.6m)= m_{T} a_{T} (1.0m)

Nf = (2.0m)(2000 kg)(9.81 m/s^{2} )/3.4m -  (0.6 m)(2320 N)/3.4m + (1.0 m)(2000 kg)(1.0 m/s^{2})  = 11541.2N - 409.4N - 588.2N = 10544N

∑fy (upward) = ∑(fy)eff: N_{f} + N_{R} - m_{T}g = 0

                                       10544 + N_{R}  - (2000kg)(9.81 m/s^{2} ) = 0

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