A 5 kg block moves in a straight line on a horizontal frictionless surface under the influence of a force that varies with posit
ion as shown in the figure. The scale of the figure's vertical axis is set by Fs = 45 N. How much work is done by the force as the block moves from the origin to x = 8.0 m?
1 answer:
The work done is 315 J
Explanation:
The graph is missing: find it in attachment.
The work done by a variable force is given by
where
F(x) is the magnitude of the force
x is the position
On a force vs position graph, the work done is equivalent to the area under the graph. Therefore, in order to find the work done as the block moves from x = 0 to x = 8.0 m, we have to calculate the area under the graph between these two points.
Each square corresponds (vertically) to 45 N, so the area of the first trapezium between x = 0 and x = 4.0 m is
Then we have to find the area of the triangle between x = 6.0 m and x = 8.0 m:
Therefore, the total area (and the work done) is
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Answer:
See the explanation below.
Explanation:
The force is a vector therefore we can decompose the force into components x & y. as we need the horizontal component of the force, we must use the cosine function of the angle.
Answer:
(a) 37.5 kg
(b) 4
Explanation:
Force, F = 150 N
kinetic friction coefficient = 0.15
(a) acceleration, a = 2.53 m/s^2
According to the newton's second law
Net force = mass x acceleration
F - friction force = m a
150 - 0.15 x m g = m a
150 = m (2.53 + 0.15 x 9.8)
m = 37.5 kg
(b) As the block moves with the constant speed so the applied force becomes the friction force.