Coyote (mc14.0 kg) is about to try out his new rocket sled invention (mr 50.0 kg

Question

Coyote (mc14.0 kg) is about to try out his new rocket sled invention (mr 50.0 kg), which can produce a thrust of 1600 N. Map Answer the three questions below. cO Part A: What is the normal force (FN) exerted upwards by the road? Number Part B: If the effective frictional coefficient between the rolling wheels and the road is 0.0700 what is the value of the frictional force (f)? Number Part C: What will be the value of the acceleration (a) while Coyote and the apparatus are in motion with the rocket firing? Number m s

Explanation / Answer

Given

Coyote mass m_c = 14 kg , mass of rocket m_r = 50 kg

thrust produced by the rocket T = 1600 N

Part (A)

the normal force exerted upward by the road is  

the weight of the system is W = m*g = (m_c+m_r) *g

N = (14+50)(9.8) = 64*9.8 N = 627.2 N

equivalent force of weight , the normal force will act on it

Part (B)

coefficient of rolling friction between wheels and road is mue_r = 0.070

the frictional force f = mue_r*N = 0.07*627.2 = 43.904 N

Part (c)

the acceleration of the rocket is  

F = thurst- frictional force

F = 1600 -627.2 N = 972.8 N

From Newton's second law F = ma

a = F/m

a = 972.8/64 m/s2

a = 15.2 m/s2

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