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1. A horizontal force of 30 N pulls a block of mass 3.6 kg across a level floor.

ID: 3893299 • Letter: 1

Question

1.A horizontal force of 30 N pulls a block of mass 3.6 kg across a level floor. The coefficient of kinetic friction between the block and the floor is


2.Consider the rock and string in the figure below. The string is fastened to a hinge that allows it to swing completely around in a vertical circle. If the rock starts at the lowest point on this circle and is given an initial speed


3.A block is dropped onto a spring with k = 29 N/m. The block has a speed of 3.5 m/s just before it strikes the spring. If the spring compresses an amount 0.10 m before bringing the block to rest, what is the mass of the block?

Consider the skateboarder in the figure below. If she has a mass of 50 kg, an initial velocity of 25 m/s, and a velocity of 13 m/s at the top of the ramp, what is the work done by friction on the skateboarder? Ignore the kinetic energy of the skateboard's wheels.

4.An electric motor is rated to have a maximum power output of 0.76 hp. If this motor is being used to lift a crate of mass 210 kg, how fast (i.e., at what speed) can it lift the crate?Hint: Assume the only other force on the crate is the force of gravity.


units in pN


Explanation / Answer

1)
net workdone, Wnet = workdone by external force - workdone by froction

Wnet = F*s - mue*m*g*s

= 30*18 - 0.25*3.6*9.8*18

= 381.24 J

we know, Wnet = chnage in kinetio cnergy

Wnet = 0.5*m*(v2^2-v1^2)


v2 = sqrt(2*Wnet/m + v1^2)

= sqrt(2*381.24/3.6 + 5^2)

= 15.4 m/s

2)

Vi sould be greater than or equal to sqrt(3*g*r)


vi = sqrt(3*g*r)

= sqrt(3*9.8*5.3)

= 12.48 m/s

3)initial mechanical energy = final mechanical ennergy

0.5*m*v^2 = 0.5*k*x^2

m = (x/v)^2*k

= (0.1/3.5)^2*29

= 0.0236 kg

= 23.6 grms

workdone by gravity + workdone friction = chnage in kinetic enrgy


m*g*(h1-h2) + Wfrivction = 0.5*m*(v2^2-v1^2)

50*9.8*(0-5) + Wfrictgion = 0.5*50*(13^2-25^2)

Wfriction = -8950 J


4)
power = F*v

v = power/F

= 0.76*743/(210*9.8)

= 0.27 m/s

5)