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A gardener pushes a 12kg lawnmower whose handle is tilted up 37 above horizontal. The lawnmower鈥檚 coefficient of rolling friction is 0.15. How much power does the gardener have to supply to push the lawnmower at a constant speed of 1.2m/s? Assume his push is parallel to the handle.

Short Answer

Expert verified

The power supplied by the gardener to push the lawnmower is 24W.

Step by step solution

01

Content Introduction

According to Newton's second law the net force acting on object is equal to its product of mass and acceleration. Therefore, the net force acting on the object is

F=ma

The weight of the force acting on the object due to gravity is given as

Fg=mg

The friction force acting on the object is

fk=N

The power required to push the object with the force F is

P=Fvcos

02

Content Explanation

Draw the free body diagram of lawnmower.

As the lawnmower is moving with constant speed, the acceleration of the lawnmower is zero which also says that net force acting on it is zero.

From the above figure, the force is acting in vertical direction N=mg+Fsin

Substitute mg+Fsinfor N in the equation fk=N

Therefore, fk=(mg+Fsin)

Refer to above figure and balance the force acting in horizontal direction,

Fcos=fk

Rearrange the equation for F,

F=mgcos-sin

The power supplied by the gardener to push the lawnmower at a constant speed v can be determined by using following equation.

P=FvcosP=(mgcos-sin)vcosP=((0.15)(12kg)(9.81m/s2)cos37-(0.15)(sin37)(1.2m/s)37P=23.87WP=24W

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Most popular questions from this chapter

A 90kg firefighter needs to climb the stairs of a 20m-tall building while carrying a 40kg backpack filled with gear. How much power does he need to reach the top in 55s?

The driver of a car traveling at 60 mph slams on the brakes, and the car skids to a halt. What happened to the kinetic energy the car had just before stopping?

A 45g bug is hovering in the air. A gust of wind exerts a force role="math" localid="1648061852575" F=(4.0i^-6.0j^)10-2N on the bug.

a. How much work is done by the wind as the bug undergoes displacement r=(2.0i^-2.0j^)m?

b. What is the bug鈥檚 speed at the end of this displacement? Assume that the speed is due entirely to the wind.

When you ride a bicycle at constant speed, nearly all the energy you expend goes into the work you do against the drag force of the air. Model a cyclist as having cross-section area 0.45m2 and, because the human body is not aerodynamically shaped, a drag coefficient of 0.90.

a. What is the cyclist鈥檚 power output while riding at a steady 7.3m/s(16mph)?

b. Metabolic power is the rate at which your body 鈥渂urns鈥 fuel to power your activities. For many activities, your body is roughly 25%efficient at converting the chemical energy of food into mechanical energy. What is the cyclist鈥檚 metabolic power while cycling at 7.3m/s?

c. The food calorie is equivalent to 4190J.How many calories does the cyclist burn if he rides over level ground at7.3m/sfor1hr?

A uniform solid bar with mass m and length L rotates with angular velocity v about an axle at one end of the bar. What is the bar鈥檚 kinetic energy?

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