Dynamics

 

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Dynamics (O Level)

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Question 1
A force of 8N is exerted on a block of 4kg. What is the acceleration of the block?
A
$2m{s^{ - 2}}$
B
$4m{s^{ - 2}}$
C
$6m{s^{ - 2}}$
D
$8m{s^{ - 2}}$
Question 1 Explanation: 
\begin{gathered} F = ma \hfill \\ 8 = 4a \hfill \\ a = 2 \hfill \\ \end{gathered}
Question 2
A block of 4kg is being pushed by a force, causing it to accelerate at $6m{s^{^{ - 2}}}$. What is the magnitude of this force?
A
0.67 N
B
1.5 N
C
2 N
D
24 N
Question 2 Explanation: 
\begin{gathered} F = ma \hfill \\ F = (4)(6) \hfill \\ a = 24 \hfill \\ \end{gathered}
Question 3
A block of 4 kg is being pushed by a force of 8 N, and experiences a frictional force of 6 N. What is the acceleration of the block?
A
$0.5m{s^{ - 2}}$
B
$0.67m{s^{ - 2}}$
C
$1.5m{s^{ - 2}}$
D
$2m{s^{ - 2}}$
Question 3 Explanation: 
Resultant force on block: $8N - 6N = 2N$ Hence, \begin{gathered} F = ma \hfill \\ 2 = 4a \hfill \\ a = 0.5 \hfill \\ \end{gathered}
Question 4
A block of 4 kg is being pushed by a force of 8 N rightwards and another force of 3 N leftwards. It also experiences a frictional force of 2 N. What is the acceleration of the block?
A
$0.5m{s^{ - 2}}$
B
$0.75m{s^{ - 2}}$
C
$1.75m{s^{ - 2}}$
D
$3m{s^{ - 2}}$
Question 4 Explanation: 
Resultant force on block: $8N - 3N - 2N = 3N$ Hence, \begin{gathered} F = ma \hfill \\ 3 = 4a \hfill \\ a = 0.75 \hfill \\ \end{gathered}
Question 5
A block of 4 kg is being pushed by a force of 8 N rightwards and another force of 3 N leftwards. It also experiences a frictional force of 5 N. What is the acceleration of the block?
A
$0m{s^{ - 2}}$
B
$0.75m{s^{ - 2}}$
C
$2m{s^{ - 2}}$
D
$2.5m{s^{ - 2}}$
Question 5 Explanation: 
Resultant force on block: $8N - 3N - 5N = 0N$ \begin{gathered} F = ma \hfill \\ 0 = 4a \hfill \\ a = 0 \hfill \\ \end{gathered}
Question 6
A ball is moving at a constant speed of $3m{s^{ - 1}}$. 2 forces of 8 N then start acting on it from directly opposite directions. Which of the following is correct?
A
The ball will stop immediately
B
The ball will slow to a stop
C
The ball will continue at its original speed
D
The ball will increase in speed
Question 6 Explanation: 
Since there is no resultant force acting on the ball, its state of motion will not be changed as well.
Question 7
2 blocks of 6 kg and 2 kg are pulled by a force of 8 N as shown. What is the acceleration of block B?
A
$1m{s^{ - 2}}$
B
$1.33m{s^{ - 2}}$
C
$3m{s^{ - 2}}$
D
$4m{s^{ - 2}}$
Question 7 Explanation: 
Since block B is attached to block A, both blocks will accelerate at the same rate due to the 8 N force. Acceleration of block A and B is: \begin{gathered} F = ma \hfill \\ 8 = (6 + 2)a \hfill \\ a = 1 \hfill \\ \end{gathered}
Question 8
2 blocks of 6 kg and 2 kg are pulled by a force of 8 N as shown. What is the tension in the string between block A and block B?
A
2 N
Hint:
6
B
4 N
C
6 N
D
8 N
Question 8 Explanation: 
Since block B is attached to block A, both blocks will accelerate at the same rate due to the 8 N force. Acceleration of block A and B is: \begin{gathered} F = ma \hfill \\ 8 = (6 + 2)a \hfill \\ a = 1 \hfill \\ \end{gathered} Since the tension on the string is the force causing block A's acceleration, \begin{gathered} F = ma \hfill \\ F = (6)(1) \hfill \\ F = 6 \hfill \\ \end{gathered}
Question 9
2 identical blocks are pulled by a force of 8 N as shown. A frictional force of 2 N acts on each of the blocks. What is the tension in the string between block A and block B? Force Pull Blocks 3
A
2 N
B
4 N
C
6 N
D
8 N
Question 10
2 blocks of 6 kg and 2 kg are pulled by a force of 8 N as shown. A frictional force of 2 N acts on each of the blocks. What is the tension in the string between block A and block B? Force Pull Blocks 2
A
3 N
B
4 N
C
5 N
D
6 N
Question 10 Explanation: 
Resultant force: \[8 - 2 - 2 = 4\] Resultant force on block A: \[4 \times \frac{6}{{6 + 2}} = 3\] Tension in string: \[3 + 2 = 5\]
Question 11
A block of 3 kg is pulling another block of 5 kg via a pulley as shown below. Assuming there is no friction, what is the acceleration of the 3 kg block? (\[g = 10m{s^{ - 2}}\]) Block Pulley Block
A
\[1.67m{s^{ - 2}}\]
B
\[3.75m{s^{ - 2}}\]
C
\[8m{s^{ - 2}}\]
D
\[10m{s^{ - 2}}\]
Question 11 Explanation: 
Force exerted by the 3 kg block: \[\begin{gathered} F = ma \hfill \\ F = (3)(10) \hfill \\ F = 10 \hfill \\ \end{gathered} \] Both the blocks accelerate uniformly. Acceleration of both the blocks: \[\begin{gathered} F = ma \hfill \\ 30 = (5 + 3)a \hfill \\ a = 3.75 \hfill \\ \end{gathered} \]
Question 12
A block of 3 kg is pulling another block of 5 kg via a pulley as shown below. Assuming there is no friction beside a frictional force of 10 N acting on the 5 kg block, what is the acceleration of the 3 kg block? (\[g = 10m{s^{ - 2}}\]) Block Pulley Block 2
A
\[2.5m{s^{ - 2}}\]
B
\[3.33m{s^{ - 2}}\]
C
\[5m{s^{ - 2}}\]
D
\[10m{s^{ - 2}}\]
Question 12 Explanation: 
Force exerted by the 3 kg block: \[\begin{gathered} F = ma \hfill \\ F = (3)(10) \hfill \\ F = 10 \hfill \\ \end{gathered} \] Resultant force on the 2 blocks: \[30 - 10 = 20\] Both the blocks accelerate uniformly. Acceleration of both the blocks: \[\begin{gathered} F = ma \hfill \\ 20 = (5 + 3)a \hfill \\ a = 2.5 \hfill \\ \end{gathered} \]
Question 13
A particle experiences 2 forces as shown below. What is the resultant force on the particle? Force on Point
A
5N, \[{36.9^o}\] East of South
B
5N, \[{36.9^o}\] West of South
C
7N, \[{36.9^o}\] East of South
D
7N, \[{36.9^o}\] West of South
Question 13 Explanation: 
Magnitude of resultant force: \[\begin{gathered} {F^2} = {4^2} + {3^3} \hfill \\ F = 5 \hfill \\ \end{gathered} \] Southwards + Westwards = Southwest direction Angle: \[\begin{gathered} \tan \theta = \frac{3}{4} \hfill \\ \theta = {36.9^o} \hfill \\ \end{gathered} \] Direction is \[{36.9^o}\] West of South
Question 14
A particle experiences 3 forces as shown below. What is the magnitude of the resultant force on the particle? Force on Point 2
A
2.1
B
2.6
C
3.2
D
4.8
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There are 14 questions to complete.

More Explainations

Qn 13

Horizontal component of diagonal force:
\[\begin{gathered}
\cos {30^o} = \frac{{{F_h}}}{3} \hfill \\
{F_h} = 2.60 \hfill \\
\end{gathered} \]

Vertical component of diagonal force:
\[\begin{gathered}
\sin {30^o} = \frac{{{F_v}}}{3} \hfill \\
{F_v} = 1.5 \hfill \\
\end{gathered} \]

Resultant horizontal force:
\[4 - 2.6 = 1.4\]
Resultant vertical force:
\[3 - 1.5 = 1.5\]
Resultant force:
\[\begin{gathered}
{F_r}^2 = {(1.4)^2} + {(1.5)^2} \hfill \\
{F_r} = 2.05 \approx 2.1 \hfill \\
\end{gathered} \]

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