Chapter 09, Problem 036 A 0.19 kg puck is initially stationary on an ice surface

Question

Chapter 09, Problem 036 A 0.19 kg puck is initially stationary on an ice surface with negligible friction. At time t = 0, a horizontal force begins to move the puck. The force is given by F = (17.2-2.381°), with F in newtons and t in its magnitude is zero. (a) what is the magnitude of the impulse on the puck from the force between t 0.547 s and t seconds, and it acts until 2.19 s7(b) what is the change in momentum of the puck between t = 0 and the instant at which F = 0? (a) Number Units (b) Number Units Click if you would like to Show Work for this question:Open Show Work

Explanation / Answer

a] The impulse is given as the integral of fdt

For the first part integrate (17.2 - 2.38t^2)dt from 0.547 to 2.19 s

The result of the integral is (17.2*t - 2.38/3* t^3).

So the impulse from 0.547 to 2.19 s is

= (17.2*2.19 - 2.38/3*2.19^3) - (17.2*0.547 - 2.38/3*0.547^3)

= 20.05

b] To find when F = 0 set 17.2 - 2.38*t^2 = 0 and solve for t

So t = sqrt(17.2/2.38) = 2.7 s

Since impulse is the change in momentum, solve the integral fdt between the limits of 0 and 2.7 s

Result = 17.2*t - 2.38/3t^3 evaluated from 0 to 2.7 s

= 17.2*2.7 - 2.38*2.7^3/3

= 30.82 Ns

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