## Physics (10th Edition)

It takes $1.2s$ for the penguin to slide to a halt.
1) The penguin slides down the incline The component of the penguin's weight along the incline $mg\sin6.9$ supports the movement downward, but kinetic friction opposes it. Since the penguin slides at a constant velocity, $$mg\sin6.9-f_k=0$$ $$1.18m-\mu_kF_N=0$$ The normal force $F_N$ balances the vertical component of the penguin's weight $mg\cos6.9$, so $F_N=mg\cos6.9=9.73m$ $$1.18m-(9.73m)\mu_k=0$$ $$1.18m=(9.73m)\mu_k$$ $$\mu_k=0.12$$ 2) The penguin enters the horizontal ice The penguin's weight no longer contributes to its horizontal movement, so only kinetic friction reduces the movement with deceleration $a$ $$f_k=-ma$$ $$(9.73m)\mu_k=-ma$$ $$a=-9.73\mu_k=-1.17m/s^2$$ The penguin start with $v_0=1.4m/s$ and its final speed is $v=0$. We find $t$: $$v=v_0+at$$ $$t=\frac{v-v_0}{a}=1.2s$$