1)when a ball is thrown vertically upwards ,its velocity goes on decreasing .what happens to its PEas its velocity becomes zero?
2)prove tht de increase in KE of a moving body is equal to the work done by de force acting on de body.
3)wht work is to be done to increase de velocity of the car from 30km/h to 60km/h if de mass of de car is 1500kg.
4)certain force acting on a 20kg mass changes it velocity from 5m/sec to 2m/sec.calculate de work done by de force??
Answers
Work = Change in kinetic energy
= 0.5 m (v² - u²)
= 0.5 × 1500 kg × [(60 × 5/18 m/s)² - (30 × 5/18 m/s)²]
= 156250 joules
= 156.25 kJ
J 1)When a ball thrown vertically upward it's velocity goes on decreasing it's potential energy increases because potential energy increases when it is raised to a height against gravity which is known as the gravitational potential energy . That's why it's kinetic energy becomes 0.
2)We know that
W=Fs where W is work done and F is force and s is displacement. Here there is a condition that initial velocity is 0. So
W=mas (F=ma)
W=m*(V2−u2)/2
W=mv2/2
This work done is stored as Kinetic Energy. So now we know that
KE=mv2/2
Now again we assume a condition where u is not 0. Following above steps we reached.
W=m*(V2−u2)/2
W=mv2/2−mu2/2
W=Final kinetic energy - Initial kinetic energy.
3)Work = Change in kinetic energy
= 0.5 m (v² - u²)
= 0.5 × 1500 kg × [(60 × 5/18 m/s)² - (30 × 5/18 m/s)²]
= 156250 joules
= 156.25 kJ
Work done is 156.25 kJ
4)Mass, m = 20 kg
Initial velocity, u = 5 m/s
Final velocity, v = 2 m/s
Time, t = 1s
From first equation of motion
v = u + at
or, a = (v - u)/t
a = (2 ms-1 - 5 ms-1)/15
a = -3 ms-2
From third equation of motion,
v² - u² = 2as
⇒ (2ms-1)² - (5ms-1)² = 2 x (-3ms-²) x s
⇒ -21 m²s-² = -6 ms-² x s
⇒ s = 7/2 m
Work done = F x s
Force = m x a
Therefore,
W = m x a x s
= 20 kg x (-3ms-²) x 7/2 m
= -210 Joules