A motor is powered by a battery with an energy capacity of E_{battery} = E battery = 1 WHr (1 WHr = 3600 Joules). The motor sits at the top of a concrete ramp and is connected to a cable that is used to pull a 68.7-kg wood crate up the concrete ramp with a coefficient of kinetic friction of 0.318 and an angle of 58.1 ^/circ ∘ above the horizontal. Assume that the net energy efficiency of the battery-powered motor is only / varepsilon = 0.25ε=0.25 for this application. (this is the fraction of the energy capacity that can be converted into useful mechanical work; the rest is lost as heat in the motor or battery). The wood crate is initially stationary at the bottom of the ramp. How far along the ramp can the motor pull the crate before the battery is completed drained? (Assume the battery is initially charged to full capacity).

Question

Physics

Aires

8 August, 11:33

A motor is powered by a battery with an energy capacity of E_{battery} = E battery = 1 WHr (1 WHr = 3600 Joules). The motor sits at the top of a concrete ramp and is connected to a cable that is used to pull a 68.7-kg wood crate up the concrete ramp with a coefficient of kinetic friction of 0.318 and an angle of 58.1 ^/circ ∘ above the horizontal. Assume that the net energy efficiency of the battery-powered motor is only / varepsilon = 0.25ε=0.25 for this application. (this is the fraction of the energy capacity that can be converted into useful mechanical work; the rest is lost as heat in the motor or battery). The wood crate is initially stationary at the bottom of the ramp. How far along the ramp can the motor pull the crate before the battery is completed drained? (Assume the battery is initially charged to full capacity).

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Landin Hurst · Answer 1

1.313m

Explanation:

Efficiency = work output / work input

0.25 = work output / 3600

Work output = 900 joules.

The load (wood crate) was on a ramp inclined at 58.1o to the horizontal;

Forces acting on the body will have vertical and horizontal component

Vertical component (the force of normal) = mgcos 58.1

Coefficient of kinetic friction = frictional force / force of normal

Frictional force = coefficient of kinetic friction * force of normal

Ff = 0.318*68.7*9.81*cos58.1

Ff = 113.25N

Horizontal force along the ramp tending to push the weight downward = mgsin58.1 = 68.7*9.81*sin58.1 = 572.16N

Total force acting downward = 572.16 + 113.25 = 685.41N

Workdone in pulling the wooden crate up = work output of the motor

Total force * distance covered = 900joules

Distance covered = 900/685.41 = 1.313m