Question

14 A force of
2i⃗ +7j⃗
N acts on a body of mass 5kg for 10 seconds. The body was initially moving with constant velocity of
i⃗ −2j⃗ m/s
. Find the final velocity of the body in m/s, in vector form.
5i⃗ +12j⃗
12i⃗ −5j⃗
10i⃗ −7j⃗
7i⃗ +10j⃗

15 The exhaust gas of a rocket is expelled at the rate of 1300 kg/s, at the velocity of 50 000 m/s. Find the thrust on the rocket in newtons
6.5×107
3.5×107
7.6×107
5.7×107

16 Sand drops at the rate of 2000 kg/min. from the bottom of a hopper onto a belt conveyor moving horizontally at 250 m/min. Determine the force needed to drive the conveyor, neglecting friction.
500 N
800 N
139 N
152 N

Answer 1

Answer on Question #58099-Physics-Other

14 A force of $2i^{+} + 7j^{+}$ N acts on a body of mass 5kg for 10 seconds. The body was initially moving with constant velocity of $i^{+} - 2j^{+} m/s$. Find the final velocity of the body in m/s, in vector form.

$$5i^{+} + 12j^{+}$$ $$12i^{+} - 5j^{+}$$ $$10i^{+} - 7j^{+}$$ $$7i^{+} + 10j^{+}$$

Solution

The acceleration vector is

$$\vec{a} = \frac{\vec{F}}{m} = \frac{(2\vec{i} + 7\vec{j})}{5} \frac{m}{s^2}.$$

The final velocity of the body is

$$\vec{v_f} = \vec{v_i} + \vec{a}t = \vec{i} - 2\vec{j} + \left(\frac{(2\vec{i} + 7\vec{j})}{5}\right)10 = (5\vec{i} + 12\vec{j}) \frac{m}{s}.$$

Answer: $5i^{+} + 12j^{+}$.

15 The exhaust gas of a rocket is expelled at the rate of $1300\ \mathrm{kg/s}$, at the velocity of $50000\ \mathrm{m/s}$. Find the thrust on the rocket in newtons

$$6.5 \times 107$$ $$3.5 \times 107$$ $$7.6 \times 107$$ $$5.7 \times 107$$

Solution

The thrust on the rocket is

$$F = \frac{dm}{dt}v = 1300\ \frac{\mathrm{kg}}{\mathrm{s}} \cdot 50000\ \frac{\mathrm{m}}{\mathrm{s}} = 6.5 \cdot 10^7\ N.$$

Answer: $6.5 \times 107$.

16 Sand drops at the rate of $2000\ \mathrm{kg/min}$ from the bottom of a hopper onto a belt conveyor moving horizontally at $250\ \mathrm{m/min}$. Determine the force needed to drive the conveyor, neglecting friction.

$$500\ \mathrm{N}$$ $$800\ \mathrm{N}$$

139 N

152 N

Solution

$$F = \frac {dm}{dt} v = 2000 \frac {\mathrm{kg\ min}}{\mathrm{min\ 60\ s}} \cdot 250 \frac {\mathrm{m\ min}}{\mathrm{min\ 60\ s}} = 139\ N.$$

Answer: 139 N.

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