Question

You throw a 3.00-N rock vertically into the air from ground level. You observe that when it is 15.0 m above the ground, it is traveling at 25.0 m/s upward. Use the work–energy theorem to find (a) the rock’s speed just as it left the ground and (b) its maximum height.

Answer 1

Answer:

(a) 30.32 m/s

(b) 46.88 m

Explanation:

Given:

Weight of the rock ($F_g$) = 3.00 N

Velocity at height 15.0 m (v) = 25 m/s

Displacement of the rock (S) = 15.0 m

(a)

Let the velocity at the start be 'u'.

The mass of the rock is given as:

$m=\frac{F_g}{g}=\frac{3\ N}{9.8\ N/kg}=0.306\ kg$

Now, the work energy theorem states that the work done by the net force is equal to the change in the kinetic energy of the body.

Here, the net force is gravitational force acting on the rock vertically downward and the displacement of the rock is vertically upward. Therefore, the work done by gravity is negative and given as:

$W_{net}=-F_g\times S =-3\ N\times 15\ m= -45\ Nm$

Now, change in kinetic energy is given as:

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

Plug in the given values and simplify. This gives,

$\Delta K=\frac{1}{2}\times 0.306(25^2-u^2)\\\\\Delta K=0.153(625-u^2)$

From work-energy theorem:

$\Delta K=W_{net}\\\\0.153(625-u^2)=-45\\\\625-u^2=\frac{-45}{0.153}\\\\u^2=625+294.12\\\\u=\sqrt{919.12}=30.32\ m/s$

Therefore, the speed with which it leaves the ground is 30.32 m/s.

(b)

Let the maximum height reached be 'h'.

We know that, at maximum height, the speed is 0 m/s.

Therefore, final speed (v) = 0 m/s

Initial speed (u) = 30.32 m/s

The work done by gravity is given as:

$W_{net}=-3h$

The change in kinetic energy is given as:

$\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}\times 0.306(0-(30.32)^2)\\\\\Delta K=0.153\times -919.12=-140.625\ J$

Again from work-energy theorem, we have:

$W_{net}=\Delta K$

$-3h=-140.625\\\\h=\frac{140.625}{3}\\\\h=46.88\ m$

Hence, the maximum height reached is 46.88 m.

Answer 2

(a) The initial speed of the rock is 30.41 m/s

(b) the maximum height of the rock is 46.25 m

Conservation of energy:

(a) The weight of the rock is W = mg = 3N

let the acceleration due to gravity be g = 10 m/s²

then the mass of the rock will be:

m = 3/g = 3/10

m = 0.3 kg

The rock is at a height of  h = 15m

the velocity of the rock at that moment, v = 25 m/s

The total energy of the ball at that height will be:

E = mgh + ¹/₂mv²

E = 3×15 + 0.5×0.3×25²

E = 138.75 J

According to the law of conservation of energy, the total energy of the rock must be conserved.

Now at the bottom potential energy is zero, so the kinetic energy is equal to the total energy.

KE = E

¹/₂mv² = 138.5 J

v² = (2×138.75) / 0.3

v² = 925 m²/s²

v = 30.41 m/s is the initial speed of the rock.

(b) from the third equation of motion:

v² = u² - 2gh

where  u = 0 will be the final speed at the maximum height

h is the maximum height

925 = 2×10×h

h = 46.25 m is the maximum height

Learn more about conservation of energy:

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