45. || a. How much work must you do to push a 10 kg block of steel across a stee

Question

45. || a. How much work must you do to push a 10 kg block of steel across a steel table at a steady speed of 1.0 m/s for 3.0 s? The coefficient of kinetic friction for steel on steel is 0.60. b. What is your power output while doing so? 46. | a. How much work does an elevator motor do to lift a 1000 kg elevator a height of 100 m at a constant speed ? b. How much power must the motor supply to do this in 50 s at constant speed? 47. ||| A 1000 kg sports car accelerates from 0 to 30 m/s in 10 s. What is the average power of the engine? 48. || In just 0.30 s, you compress a spring (spring constant 5000 N/m), which is initially at its equilibrium length, by 4.0 cm. What is your average power output? 49. || An elite Tour de France cyclist can maintain an output power of 450 W during a sustained climb. At this output power, how long would it take an 85 kg cyclist (including the mass of his bike) to climb the famed 1100-m-high Alpe d'Huez mountain stage? 50. || A 710 kg car drives at a constant speed of 23 m/s. It is subject to a drag force of 500 N. What power is required from the car's engine to drive the car a. On level ground? b. Up a hill with a slope of 2.0 degree ? 51. ||| An elevator weighing 2500 N ascends at a constant speed of 8.0 m/s. How much power must the motor supply to do this?

Explanation / Answer

45). Mass of steel block, m= 10kg

Steady speed of block, u= 1.0m/s

Time taken, t= 3.0s

Then distance covered in above time will be d= u*t= (1)(3.0)= 3.0m

Now since there is no motion of block in perpendicular direction of the motion thus normal reaction force on the block by surface will be

R= mg= (10)(9.8)= 98 N

Then frictional force on the block will be,

F= uR

where u= coefficient of friction= 0.60

Thus frictional force on the block will be, F= (0.60)(98)= 58.8 N

Then work done by pushing will be

W= F.d= Fdcos180= -Fd (since frictional force acts in opposite direction of the motion)

using all given values in above,

W= -(58.8)(3.0)=-176.4J (ANS)

b). Power output while doing so will be

P= |W|/t

using all given and calculated values in above,

P= |-176.4|/3.0= 58.8 W (ANS)

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