Numerical - Work done - Work energy principle - Mechanics

Q.  A block is displaced by 3 m when a force of 200 N is applied on it on an inclined surface which is at an angle of 50o with the horizontal. What is the work done?
- Published on 21 Sep 15

a. 385.67 Nm
b. 459.62 Nm
c. 933.00 Nm
d. Insufficient data

ANSWER: 385.67 Nm

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    Discussion

  • Osaid Ahmad   -Posted on 20 Jan 21
    It is important to mention in the question that force applied is horizontal
  • Arslan Liaqat   -Posted on 17 Aug 20
    Work done = Force x cos ? x D

    = 200 x cos 50 x 3
    = 579J
  • Sravanthi   -Posted on 15 Dec 15
    Given: Force = 200 N, θ = 50°, block displacement (D) = 3m

    Formula: Work done = Force x cos θ x D

    Solution:

    Work done = Force x cos θ x D

    = 200 x cos 50 x 3

    = 385.67 Nm

    Work done = 385.67 Nm

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