Area under the graph = impulse and is similar enough (less than 25% error) to be able to calculate the impulse of the curved graph.
A=(1/2)(b)(h)
A=(1/2)(0.062s-0.012s)(35N)
A=0.875N*s
The force shown in the figure below is the net eastward force acting on a ball. The force starts rising at t=0.012 s, falls back to zero at t=0.062 s, and reaches a maximum force of 35 N at the peak. Determine with an error no bigger than 25% (high or low) the magnitude of the impulse (in N-s) delivered to the ball.
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