A factory worker pushes a 32.0 kg crate a distance of 7.0 m along a level floor at constant velocity by pushing horizontally on it. The coef

Question

A factory worker pushes a 32.0 kg crate a distance of 7.0 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and floor is 0.26.

Required:
a. What magnitude of force must the worker apply?
b. How much work is done on the crate by this force?
c. How much work is done on the crate by friction?
d. How much work is done on the crate by the normal force? By gravity?
e. What is the total work done on the crate?

in progress 0
Eirian 2 weeks 2021-07-19T14:04:38+00:00 1 Answers 5 views 0

Answers ( )

    0
    2021-07-19T14:05:44+00:00

    Answer:

    (a) 81.54 N

    (b) 570.75 J

    (c) – 570.75 J

    (d) 0 J, 0 J

    (e) 0 J  

    Explanation:

    mass of crate, m = 32 kg

    distance, s = 7 m

    coefficient of friction = 0.26

    (a) As it is moving with constant velocity so the force applied is equal to the friction force.

    F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N

    (b) The work done on the crate

    W = F x s = 81.54 x 7 = 570.75 J

    (c) Work done by the friction

    W’ = – W = – 570.75 J

    (d) Work done by the normal force

    W” = m g cos 90 = 0 J

    Work done by the gravity

    Wg = m g cos 90 = 0 J

    (e) The total work done is

    Wnet = W + W’ + W” + Wg = 570.75 – 570.75 + 0 = 0 J  

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )