An athlete throws the shot-put of mass 4 kg with initial speed of 2.2 ms-1 at 41° from a height of 1.3 m from the ground. What is the KE of the shot-put when it reaches the ground? (Ignoring the air resistance and gravity g=9.8 m/s-2 )

Question

An athlete throws the shot-put of mass 4 kg with initial speed of 2.2 ms-1 at 41° from a height of 1.3 m from the ground. What is the KE of the shot-put when it reaches the ground? (Ignoring the air resistance and gravity g=9.8 m/s-2 ) (A)  42.84 J  (B)  52.84 J  (C)  62.84 J  (D)  72.84 J

Tardigrade Admin · Accepted Answer

As there is no air resistance and gravitational force is a conservative force, we can apply mechanical energy conservation.

\Rightarrow (K E)_{f} = (K E)_{i} + Change in PE

= \frac{1}{2} m V^{2} + m g h

= \frac{1}{2} \times 4 \times (2.2)^{2} + 4 \times (9.8) \times (1.3)

= 9.68 + 50.96

= 60.64 \approx 62.84 J

Q. An athlete throws the shot-put of mass $4 k g$ with initial speed of $2.2 m s^{- 1}$ at $4 1^{\circ}$ from a height of $1.3 m$ from the ground. What is the KE of the shot-put when it reaches the ground? (Ignoring the air resistance and gravity $g = 9.8 m / s^{- 2}$ )

$42.84 J$

$52.84 J$

$62.84 J$

$72.84 J$

Q. An athlete throws the shot-put of mass 4kg with initial speed of 2.2ms−1 at 41∘ from a height of 1.3m from the ground. What is the KE of the shot-put when it reaches the ground? (Ignoring the air resistance and gravity g=9.8m/s−2 )

42.84J

52.84J

62.84J

72.84J

As there is no air resistance and gravitational force is a conservative force, we can apply mechanical energy conservation. ⇒(KE)f​=(KE)i​+ Change in PE =21​mV2+mgh =21​×4×(2.2)2+4×(9.8)×(1.3) =9.68+50.96 =60.64≈62.84J

Q. An athlete throws the shot-put of mass $4 k g$ with initial speed of $2.2 m s^{- 1}$ at $4 1^{\circ}$ from a height of $1.3 m$ from the ground. What is the KE of the shot-put when it reaches the ground? (Ignoring the air resistance and gravity $g = 9.8 m / s^{- 2}$ )

$42.84 J$

$52.84 J$

$62.84 J$

$72.84 J$