Answer to Question #150381 in Classical Mechanics for Jacob stan

Question #150381
A child starts sledding from rest, going down a 40-m-long 7.5° incline, then coasting across a horizontal stretch. The mass of sled + child is 35 kg, and the coefficient of kinetic friction is 0.060.
What is the total time for the ride?
1
Expert's answer
2020-12-14T07:24:59-0500

inclined track:\text {inclined track:}

F1=ma1F_1=ma_1

a1=g(sinθμcosθ)a_1=g(\sin \theta -\mu \cos \theta )

a1=9.8(sin7.500.06cos7.50)=0.7a_1 = 9.8(sin7.5^0-0.06*cos 7.5^0) = 0.7

s=V0t1+a1t122;V0=0s = V_0t_1+\frac{a_1t_1^2}{2};V_0=0

t1=2sa1=2400.7=10.7t_1 = \sqrt{\frac{2s}{a_1}}=\sqrt{\frac{2*40}{0.7}}=10.7

V1=V0+at1V_1 = V_0+at_1

V1=0.710.7=7.49V_1=0.7*10.7 =7.49


horizontal track:\text{horizontal track:}

F2=ma2F_2=ma_2

a2=μga_2= \mu*g

V2=V0at2V_2= V_0-at_2

V2=0 stop motionV_2=0 \text{ stop motion}

V0=7.49 speed at the end inclined trackV_0 = 7.49 \text{ speed at the end inclined track}

t2=V0a2=V0μg=7.499.80.06=12.7t_2= \frac{V_0}{a_2}=\frac{V_0}{\mu*g}=\frac{7.49}{9.8*0.06}=12.7


All time:\text{All time:}

t=t1+t2=12.7+10.7=23.4t = t_1+t_2 = 12.7+10.7 = 23.4

Answer:total time 23.4 seconds






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