If the rocket sled shown in the figure below starts with only one rocket burning

Question

If the rocket sled shown in the figure below starts with only one rocket burning, what is the magnitude of its acceleration (in m/s2)? Assume that the mass of the system is 2140 kg, the thrust T is 1.1 104 N, and the force of friction opposing the motion is known to be 790 N.

________________m/s2

(b) Why is the acceleration not one fourth of what it is with all rockets burning? (The acceleration when all four rockets are burning is 20 m/s2. The force of friction remains 790 N.)

Question 2

Alex is asked to move two boxes of books in contact with each other and resting on a frictionless floor. He decides to move them at the same time by pushing on box A with a horizontal pushing force

p = 8.6 N. Here box A has a mass mA = 12.1 kg and box B has a mass mB = 7.0 kg. The contact force between the two boxes is

c and

denotes the normal force.

(a) What is the acceleration of the two boxes?
_______m/s2

(b) What is the force exerted on mB by mA?
__________N

(c) If Alex were to push from the other side on the 7.0 kg box, how would your answer to part (b) change?
_________N

Explanation / Answer

Q1) a)

T - f = m*a

1.1*10^4 - 790 = 2140*a

11000 - 790 = 2140*a

a = 4.771 m/s^2

b) It is not one fourth of the acceleration when all the rockets operate because:-

when all the 4 rockets work

4T - f' = m*a' (f' is friction and a' is acceleration when all the rockets work)

also, T - f = m*a

now for a' = 4a

i.e 4T - f' = 4ma

f' = 4(T - ma)

f' = 4f

i.e the frictional force(when all the 4 rockets work) should also be 4 times of the friction when only 1 rocket is working. Since the friction remains the same, the acceleration isnot exactly one fourth.

Q2)a) Fp = (m1+m2)a

8.6 = (12.1+7.0)*a

a = 0.4502 m/s^2

b) acceleration of both the boxes is same

so, the force exerted on Mb by Ma (Fc) = Mb*a

Fc = 7.0*0.4502

Fc = 3.151 N

c) If he pushes from the other side, the acceleration will be the same, i.e

a = 0.4502 m/s^2

so the force on Ma by Mb = force on Mb by Ma = contact force = Fc = Ma*a

Fc = 12.1*0.4502

Fc = 5.44 N

The force applied on Mb by Ma is 5.44 N

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