## College Physics (4th Edition)

(a) The crate travels $17.6~m$ down the incline in $2.50~s$ (b) The crate's speed after $2.50~s$ is $14.1~m/s$
(a) We can find the crate's acceleration: $mg~sin~\theta = ma$ $a = g~sin~\theta$ $a = (9.80~m/s^2)~sin~35.0^{\circ}$ $a = 5.62~m/s^2$ We can find the distance the crate travels down the incline in $2.50~s$: $d = \frac{1}{2}at^2$ $d = \frac{1}{2}(5.62~m/s^2)(2.50~s)^2$ $d = 17.6~m$ The crate travels $17.6~m$ down the incline in $2.50~s$ (b) We can find he speed after $2.50~s$: $v_f = v_0+at$ $v_f = 0+(5.62~m/s^2)(2.50~s)$ $v_f = 14.1~m/s$ The crate's speed after $2.50~s$ is $14.1~m/s$.