Question

3 one please please please​

Answer 1

Question 3.

What is the work done to change the velocity of an object from 76km/h to 36 km/h if its mass is 1500kg? Does the work done by the force negative or positive?

Solution:

• Initial velocity (u) = 76 km/h

Converting into m/s

So, u = 76 * (5/18) = (380)/18 = 21.1 m/s

• Final velocity (v) = 36 km/h

Converting into m/s

So, v = 36 * (5/18) = 2 * 5 = 10 m/s

Mass of the object (m) = 1500kg

Now,

Work Done = $\frac{1}{2}mv^{2} - \frac{1}{2}mu^{2}$

$= \frac{1}{2}m (v^{2} - u^{2})$

By taking ('1/2m' as common)

Now putting the values, we get,

$= \frac{1}{2} * 1500 \big[ (10)^{2}-(21.1)^{2}\big]$

$= \frac{1}{2} * 1500 \big[ (100)-(445.21)\big]$

$= \frac{1}{2} * 1500 \big[ (-345.21)\big]$

$= 750 * (-345.21)$

$=\boxed{\boxed{\bf - 258907.5\ J}}$

Hence,

The work done by the force is negative.

.

Question 4.

An object of mass 2kg raised to a height 10m possess potential energy of 200J. What is the kinetic energy and potential energy at a height 4 metre?

Solution:

Mass of the object (m) = 2kg

Now,

The potential energy = mgh

For height 4m,

PE = 2 * 10 * 4

PE = 80 J

• Here no force like friction, so we can use
• ∆U = ∆K

Intital PE + initial KE = final PE + final KE

Then

→ mg(10) + 0 = mg(4) + KE

→ 2 * 10 * 10 = 2 * 10 *4 + KE

→ 200 = 80 + KE

→ 200 - 80 = KE

→ 120J = KE

Hence,

PE (Potential Energy) at 4m height = 80J

And

KE (Kinetic Energy) at 4m height = 120J