VU Answer Provide GSC101 Assignment 1 Solution Fall 2021. Easy to See Step by Step Correct GSC101 Assignment 1 Solution 2021 Download PDF File Below.

GSC101 ASSIGNMENT 1 SOLUTION FALL 2021

Provide by VU Answer

Total Marks: 20

Due Date: 12 Dec 2021

Question 1:

Write down two real-life examples where the object is moving with positive velocity but having negative acceleration?

Solution:

1. Step on the breaks of your car you have positive velocity and negative acceleration assuming that your coordinate system is situated such that the direction you are traveling is positive.

2. Stone threw straight up. As long as it is climbing up, it has positive velocity and negative acceleration.

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Question 2:

Read the following statements carefully, discuss which Newton's law of motion is applicable, and also explain the reason.

Solution:

Question 3:

In the given figure two objects are hung on a frictionless pulley. The block of mass (m1) moves down and the bucket of mass (m2) moves up. Find out the magnitude of acceleration (a) of these two objects and tension (T) in the string. (Neglect the masses of pulley and rope)

Solution:

The force that affects the bucket m2:

W2 --------weight

T---------------Tension force by which the rope affects the bucket

The force that affects the block m1:

W1-------weight

T1 -------Tension force by which the rope affect the block

T2 -------Tension of force by which the rope affect the block

For bucket the equation to scalar form:

T-W2=m2a2--------1

W1-T1-T2=m1a1-------2

Because we leave the mass of the pulley system and the rope out of the account, they have no moment of inertia and don’t affect the tension of the forces. The following hold for the magnitude of the tension force.

T=T1=T2

We write the above equation

T-W2 =m2 a2 ------ 3

W1-2T=m1a1 ------- 4

Relation between the magnitude of the acceleration a1 and a2

If the bucket goes up a distance S, the block goes down a distance S/2

S=a2 t2

=a1 t2

By above two-equation to get

2=

2a1=a2 ----- 5

So, the equation (3) and equation (4) can be written as

T-m2 g=m22a1 ------- 6

m1g - 2T=m1a1 ------7

We have two equations (6) and (7) in two variables T and a1.

From them we can determine the magnitude of the acceleration a1 and the force T. we multiply equation (6) by 2 and both equations up:

m1g+2m2g=m1a1+4m2a1

(m1+2m2)g= (m1+4m2)a1

a1 = ------ 8

According to equation 5

a2= 2------ 9

The magnitude of the tension force T is given for example by the equation (6)

T-m2g=m22a1

T=m22a1+m2g

T= m22+m2g

After simplify we get

T=--------10

Put m1=30 kg, m2=15 kg and g=10m/s2 in equation (8) (9) and (10)

a1=

a1==6.67m/s2

Similarly

a2 = 2(6.67) =13.33m/s2

And

T= = 150N

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