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Gnesinka [82]
3 years ago
8

A horizontal force of 40N is needed to pull a 60kg box across the horizontal floor at which coefficient of friction between floo

r and box? Determine it to three significant figures even through that's quite unrealistic. How much work is done in overcoming friction between the object and floor if the box slides 8m along horizontally on the floor?
Physics
1 answer:
Mazyrski [523]3 years ago
8 0

Answer:

Coefficient of friction is 0.068.

Work done is 320~J.

Explanation:

Given:

Mass of the box (m): 60 kg

Force needed (F): 40 N

The formula to calculate the coefficient of friction between the floor and the box is given by

F=\mu mg...................(1)

Here, \mu is the coefficient of friction and g is the acceleration due to gravity.

Substitute 40 N for F, 60 kg for m and 9.80 m/s² for g into equation (1) and solve to calculate the value of the coefficient of friction.

40 N=\mu\times60 kg\times9.80 m/s^{2} \\~~~~~\mu=\frac{40 N}{60 kg\times9.80 m/s^{2}}\\~~~~~~~=0.068

The formula to calculate the work done in overcoming the friction is given by

W=Fd..........................(2)

Here, W is the work done and d is the distance travelled.

Substitute  40 N for F and 8 m for d into equation (2) to calculate the work done.

W=40~N\times8~m\\~~~~= 320~J

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A small mailbag is released from a helicopter that is descending steadily at 3 m/s.
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<u>Answer:</u>

a) Speed of mailbag after 3 seconds = 32.4 m/s

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    Distance traveled by helicopter = 9 meter.

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  s= 3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 53.1m

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c) Initial velocity = -3 m/s, acceleration = 9.8 m/s^2 and time = 3 seconds.

  v = -3+9.8*3 = 26.4 m/s

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    s= -3*3+\frac{1}{2} *0*3^2\\ \\ s=-9m

    Distance traveled by helicopter = 9 meter.

 Velocity of package = -3 m/s, time taken = 3 seconds, acceleration = 9.8 m/s^2.

  s= -3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 35.1m

  Distance traveled by package  = 35.1 meter.

So package is (35.1+9)meter below helicopter = 44.1 m

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