Question

Question. No: 1
A lift is filled with patients has a total mass of 2055 kg. As the lift begins to go up, the acceleration is 0.75 m/s2. What is the tension in the rope that is lifting the lift? Validate your solution as well.
Question. No: 2
A man pushes a lawnmower with a force of 170 N at an angle of 37° down from the horizontal. The lawn is 12.0 m wide and requires 16 complete trips across and back. How much work does he do?
Assignment 2: (Spring 2014)
Question. No: 3
A bus’s tire rotates at an initial angular speed of 20.5 rad/s. The driver accelerates, and after 4.5 s the tire’s angular speed is 29.0 rad/s. What is the tire’s average angular acceleration during the 4.5 s time interval?
Question. No: 4
An object’s momentum depends upon the object’s
a) mass, speed and acceleration.
b) mass, speed and direction of motion.
c) speed and acceleration.
d) velocity and direction of motion.
e) mass and acceleration.
Write the reason of your above selected choice as well.
Question. No: 4
You drive a nail horiz

Answer 1

Answer on Question #42811-Physics-Mechanics | Kinematics | Dynamics

Question. No: 1

A lift is filled with patients has a total mass of $m = 2055 \, kg$. As the lift begins to go up, the acceleration is $a = 0.75 \frac{m}{s^2}$. What is the tension in the rope $T$ that is lifting the lift?

Solution

We can apply Newton's Second Law:

$$m a = T - m g \rightarrow T = m (a + g) = 2 0 5 5 k g \left(0. 7 5 \frac {m}{s ^ {2}} + 9. 8 1 \frac {m}{s ^ {2}}\right) = 2 1. 7 k N.$$

Answer: 21.7 kN.

Question. No: 2

A man pushes a lawnmower with a force of $F = 170 \, N$ at an angle of $37{}^{\circ}$ down from the horizontal. The lawn is $d = 12.0 \, m$ wide and requires 16 complete trips across and back. How much work does he do?

Solution

The work is

$$W = 1 6 \cdot F d \cos 3 7 {}^ {\circ} = 1 6 \cdot 1 7 0 N \cdot 1 2. 0 m \cdot \cos 3 7 {}^ {\circ} = 2 6. 1 k J.$$

Answer: 26.1 kJ.

Question. No: 3

A bus's tire rotates at an initial angular speed of $\omega_{i} = 20.5\frac{rad}{s}$. The driver accelerates, and after 4.5 s the tire's angular speed is $\omega_{f} = 29.0\frac{rad}{s}$. What is the tire's average angular acceleration during the 4.5 s time interval?

Solution

The tire's average angular acceleration during the $t = 4.5$ s time interval is

$$\alpha = \frac {\omega_ {f} - \omega_ {i}}{t} = \frac {2 9 . 0 \frac {r a d}{s} - 2 0 . 5 \frac {r a d}{s}}{4 . 5 s} = 1. 8 9 \frac {r a d}{s ^ {2}}.$$

Answer: .89 $\frac{rad}{s^2}$.

Question. No: 4

An object's momentum depends upon the object's

a) mass, speed and acceleration.

b) mass, speed and direction of motion.

c) speed and acceleration.

d) velocity and direction of motion.

e) mass and acceleration.

Solution

An object's momentum $\vec{P}$ is

$$\vec{P} = m \vec{v}.$$

Therefore, it depends upon the object's mass, speed and direction of motion.

**Answer:** b) mass, speed and direction of motion.

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