Lee pushes horizontally with a force of 88 N on a 12 kg mass for 11 m across a f

Question

Lee pushes horizontally with a force of 88 N on a 12 kg mass for 11 m across a floor. Calculate the amount of work Lee did. Answer in units of J A man cleaning his apartment pushes a vacuum cleaner with a force of magnitude 45.6 N. The force makes an angle of 22.5 degree with the horizontal floor. The vacuum cleaner is pushed 3.9 m to the right along the floor. Calculate the work done by the 45.6 N force. Answer in units of J A force F is exerted by a broom handle on the head of the broom, which has a mass m. The handle is at an angle theta to the horizontal, as shown below. What is the work done by the force on the head of the broom as it moves a distance d across a horizontal floor? W = Fdsin theta W = Fmcos theta W = Fmtan theta W = Fmdsinn theta Find the work done by the dogs. Answer in units of kJ Find the energy lost due to friction. Answer in units of kJ If 3.7 J of work is done in raising a 178 g apple, how far is it lifted? Answer in units of rn Starting from rest, a 4.58 kg block slides 2.99 m down a rough 24.1 degree incline. The coefficient of kinetic friction between the block and the incline is 0.393. The acceleration of gravity is 9.8 m/s2. Find the work done by the force of gravity. Answer in units of J Find the work done by the friction force between block and incline. Answer in units of J Find the work done by the normal force. Answer in units of J A cheerleader lifts his 29.1 kg partnerr straight up off the ground a distance of 0.927 m before releasing her. The acceleration of gravity is 9.8 rn/s2. If he does this 20 times, how much work has he done? Answer in units of J

Explanation / Answer

ans 11) work = force * displacement = 88*11 =968

ans 12 )work = force * displacement = f*d cos(theta) = 164.30

ans13 ) work = force * displacement = f*d cos(theta) {none of the options are correct}

ans14 ) no question

ans15) no

ans 16) work =3.7

force = 0.178*9.8

d = W/F = 2.11 metre

ans 17 ) component of gravitational force in the direction of displacement = m* g sin(24.1) = 4.58*9.8* sin(24.1 ) = 18.32 N

work = 18.32* 2.99 = 54.77 J

ans 18) friction force = mu *N

Normal component (N) = m*g*cos(24.1)

mu = 0.393 (coefficient of kinetic friction)

friction force = 16.10

work = force * displacement = 16.10 *2.99 = 48.139 J

ans 19 ) work done by normal force = 0

since no displacement takes place in the direction of force

ans 20)since the work depends only on final displacement it wont depend on number of time she pick him up

work = 29.1 *9.8*0.927 = 264.36 J

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