10.
A horizontal force F is used to pull a box placed on a floor. Variation in the force with position coordinat
x measured along the floor is shown in the graph.
F (N)
X
(m)
5
10 T
(a) Calculate the work done by the force in moving the box from x=0 to x=10 m.
(b) Calculate the work done by the force in moving the box from x=10 m to x=15 m.
(c) Calculate the work done by the force in moving the box from x=0 to x=15 m.
Answers
Ans
(A) 75J (B) -25J (C) 50J
Explanation
Work =Area between the curve and displacement axis
So,
(A) W= 10×5+1/2×5×10 x=0m to x=10m
=50+25=75J
(B) W=1/2×5×-(10) x=10m to x=15m
=-25J
(C) W=75-25=50J x=0m to x=15m
Answer:
a) 75 J
b) -25 J
c) 50J
Explanation:
You have to plot the graph F(N) in y-axis and m in y-axis
Take the values from the graph
Two angle will be found one on the positive side of x-axis and the other on the negative side of the y-axis.
dW = f. dx
Integrate on both the sides we get
∫ dW = ∫₀¹⁰ f dx ⇒ area under the graph from 0 to 10
Area under the curve gives the work done
a) To calculate the work done by the force in moving the box from x=0 to x=10 m
Area = ((1/2) × 5×10) + (5×10)
= 25 + 50
= 75 J
Work done = 75 J
b) To calculate the work done by the force in moving the box from x=10m to x=15 m
Area = (1/2) × (-10) × 5
= -25J
work done = -25 J
c) To calculate the work done by the force in moving the box from x=0 to x=15 m
Note:To find the entire work done from x=0 to x= 15 m add the work done for x=0 to x=10m and x=10m to x= 15m
Work done = 75 + (-25)
= 50
Work done = 50 J
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