Hello students welcome to new post. In this post we will discuss Fsc 1st year physics chapter 4 solved exercise. We will learn all exercise question of Fsc part one physics chapter 4. All questions will be discuss with the details. In previous tutorial we has discuss chapter 3 exercise short question. So let get started
Fsc 1st year physics chapter 4 solved exercise
Q 4.1: Solution
- In these two conditions there is no work done since value of displacement for both objects or bodies is zero.
- As we know
- w= F.d
- =Fdcosθ
- as value of displacement is zero so
- W=F(0)Cosθ
- so Work done is zero
Q 4.2: Solution
- As we given
- m=10 kg
- h= 10 m
- He have to find work done in kilojoules
- Work done =w=mgh
- putting values
- W= 10 x 10x 9.8
- W=980 J
- W=(980 x1000)/1000
- 1000= 1 Kilo
- 980/1000 kJ
Q 4.3: Solution
- To understand this problem diagram is created here you can see.
- Due to to Force F distance L covered so work done is FL Accordioning to our condition distance is 2L due to force of 3F. In results
- W= 3F.2L
- 6FL
- So total work done will be
- W=Fl+ 6FL= 7FL
Q 4.4: Solution
- For first condition
- m=50 kg
- h=50 cm=0.5m
- Value of wok for this condition
- Wx= mgh
- =50 x0.5 x 9.8
- =245 J
- For second condition
- m=50kg
- d=2m
- F=50 N
- Value of work for this condition is
- Wy=F.d
- 50 x 2
- =100 J
- More work done in first condition
Q 4.5: Solution
- As we know one Joule = 1 N x 1 m
- If one N force is applied on object to to move it one meter height. In results work done stores in the object in form of P.E that is one joule body has ability to perform one joule work
Q 4.6: Solution
- Ball is at height of h1 from table and table at h2 height from ground. According to ground height of ball is h1+h2.
- Potential energy stored in the ball is mgh1 and mg(h1+h2). Both students are correct one is taking table as first point and other ground.
Q 4.7 Solution
- When rocket re-enters into the atmospheres then its some K.E uses against the air friction and agins dust particles that exist in form of heat that causes turn nose of racket hot
Q 4.8 Solution
- a: Compressed spring has elastic P.E
- b: Water in high dam has gravitational p.E
- c: Moving car has K.E
Q 4.8 Solution
- For curtain height cup has gravitational P.E. When cup drops down its gravitational P.E reduces and K.E rises. Before colliding to the grounds its K.E has extreme value and after colliding converting to sound energy. With that heat energy and in energy that used to break the cup in case of air friction is not counted.
- If friction is exist than some portion of K.E is used to reduce the friction value.
Q 4.9 Solution
- Elastic potential energy stored in the catapult is transferred to the stone in form of K.E
- When stone collide to the window its K.E converted into sound, heat and energy used for breaking the window.
That is all about the Fsc 1st year physics chapter 4 solved exercise. I have explained all questions of chapter 4. I hope that will helps you to understand and resolve your queries. In next post we will discuss chapter 5 question till than have a nice day.
