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**GSC101 Assignment No 1 Solution Fall 2023 – VU Assignment Solution – VU Fall Assignment Solution**

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**Assignment No. 1 **

**Solution ****Fall 2023**

**GSC101 Assignment 1**

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**Question-1: (Marks=4+4=8)**

**Two blocks, denoted as m1 and m2, are linked by a slender string that runs over a pulley, as depicted in the illustration. Blocks**

are held at rest and then released

. If the coefficient of friction between block 1 and table surface is Î¼= 0.5,

(a). Find the acceleration (a) of the system.

(b). Determine the tension (T) in the string.

**Solution:**

Given data:

â€¢ Mass of block 1 (1m1): 10 kg

â€¢ Mass of block 2 (m2): 15 kg

â€¢ Coefficient of friction (Î¼): 0.5

Acceleration (a) and Tension (T) are the unknowns.

Step 1: Set up the Equations of Motion

For Block 1 (m1):

m1â‹…gâˆ’Tâˆ’Î¼â‹…m1â‹…g=m1â‹…a

For Block 2 (m2):

Tâˆ’m2â‹…g=m2â‹…a

Step 2: Substitute Known Values

Given =9.8 m/s2g=9.8m/s2:

For Block 1:

10â‹…9.8âˆ’Tâˆ’0.5â‹…10â‹…9.8=10â‹…a

For Block 2:

Tâˆ’15â‹…9.8=15â‹…a

Step 3: Solve the System of Equations

Now, we need to solve these equations simultaneously to find a and T.

Step 4: Solve for a

From the first equation: 98âˆ’Tâˆ’49=10a

âˆ’T=10aâˆ’49

From the second equation: T-15a+147

Now, substitute the expression for T into the equation âˆ’T=10aâˆ’49 and solve for a:

âˆ’49âˆ’(15a+147)=10aâˆ’

15aâˆ’147=10aâˆ’49

âˆ’25-98

a=âˆ’25/98

Step 5: Solve for T

Now substitute the value of a back into the expression for T:

=15â‹…(âˆ’25/98)+147

Step 6: Numerical Calculation

T-15â‹…(âˆ’98/25)+147

Now calculate a and T.

Results:

Aâ‰ˆâˆ’3.92m/s2

Tâ‰ˆ102.12N

Explanation:

The negative sign for a indicates that the acceleration is upward, opposite to the direction assumed.

The tension (T) in the string is approximately 102.12N.

**Question 2: (Marks=4+4=8)**

**Consider two cars positioned on a linear roadway. Initially, Car A is stationary, whereas Car B maintains a constant velocity. At**

a specific point in time, Car A begins to accelerate, Determine the following:

(a). The time required for Car A to reach the same position as Car B.

(b). The displacement of both Car A and Car B at the moment they come together.

**Solution:**

Given data:

â€¢ Initial velocity of Car A (0VA0): 0 m/s (as it is initially stationary)

â€¢ Initial velocity of Car B (0VB0): 1010 m/s

â€¢ Acceleration of Car A (aA): 44 m/s22

â€¢ Time (t): unknown

â€¢ Displacement of both cars at the moment they come together (s): unknown

Step 1: Find the Time (t)

The equation relating initial velocity, acceleration, and time for uniformly accelerated motion is:

VA=VA0+aAâ‹…t

Substitute the given values for Car A:

0+4=100+4t=10

Solve for t:

t-10/4-2.5s

Step 2: Find the Displacement (s)

Now that we know t, we can find the displacement using the kinematic equation:

2sA=VA0â‹…t+21â‹…aAâ‹…t2

Substitute the known values:

=0â‹…2.5+21â‹…4â‹…(2.5)2

=0+2â‹…6.25

=12.5m

Since Car B has a constant velocity, its displacement is given by:

sB=VB0â‹…t

Substitute the values for Car B:

SB =10â‹…2.5

SB=25m

Results:

(a) The time required for Car A to reach the same position as Car B is 2.52.5 seconds.

(b) The displacement of Car A is 12.512.5 meters, and the displacement of Car B is 2525 meters at the moment they

come together.

**Question 3: (Marks=4)**

**Provide two real-world scenarios in which an object can exhibit acceleration while maintaining a velocity of zero.**

**Answer:**

**Elevator Scenario:**

Imagine you are standing in an elevator that starts moving upward. When the elevator begins to accelerate, your

velocity is initially zero because you were at rest. However, as the elevator accelerates, you experience a change in

velocity. The acceleration occurs even though your initial velocity was zero. This scenario is an example of zero

initial velocity with upward acceleration.

Car at a Traffic Light:

Consider a car at a traffic light. When the traffic light turns green, the driver presses the accelerator, causing the car

to accelerate. Initially, the car is at rest (velocity is zero), but due to the acceleration provided by the engine, the car

starts moving forward. During this period of acceleration, the velocity increases while initially being zero. However,

the car maintains a velocity of zero at rest until the driver accelerates.

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