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:
h = 3.5 m
Explanation:
First, we will calculate the final speed of the ball when it collides with a seesaw. Using the third equation of motion:
where,
g = acceleration due to gravity = 9.81 m/s²
h = height = 3.5 m
vf = final speed = ?
vi = initial speed = 0 m/s
Therefore,
Now, we will apply the law of conservation of momentum:
where,
m₁ = mass of colliding ball = 3.6 kg
m₂ = mass of ball on the other end = 3.6 kg
v₁ = vf = final velocity of ball while collision = 8.3 m/s
v₂ = vi = initial velocity of other end ball = ?
Therefore,
Now, we again use the third equation of motion for the upward motion of the ball:
where,
g = acceleration due to gravity = -9.81 m/s² (negative for upward motion)
h = height = ?
vf = final speed = 0 m/s
vi = initial speed = 8.3 m/s
Therefore,
<u>h = 3.5 m</u>