Question

A truck has four times the mass of a car and is moving with twice the speed of the car. If Kt and Kc refer to the kinetic energies of truck and car respectively, it is correct to say that:A) Kt = 16Kc. B) Kt = 4Kc. C) Kt = Kc. D) Kt = 2Kc.

Answer 1

Answer:

A) Kt = 16Kc.

Explanation:

Kinetic Energy: This is the energy of a body in motion. The S.I unit of kinetic energy is Joules (J).

Mathematically kinetic energy can be represented as

K = 1/2mv².............................. Equation 1

Where K = kinetic Energy, m = mass of the body, v = velocity of the body.

(i) The kinetic energy (Kc) of thevcar,

Kc = 1/2m₁v₁²

Where m₁ = mass of the car, v₁ = velocity of the car.

Kc = 1/2(m₁v₁)

If the mass of the truck is four times the mass of the car and is moving with twice the speed of the car.

Therefore,

Kt = 1/2(4m₁)(2v₁)²

Kt = 8m₁v₁²

From the above,

Kt/Kc = 4m₁v₁²/(1/2m₁v₁²)

Kt/Kc = 16

Kt = 16Kc

Thus the right option is A) Kt = 16Kc.

Answer 2

The relationship between the kinetic energy of truck and the car is $K_t = 16K_c$.

The given parameters;

• mass of the car, = $m_c$
• mass of the truck, $m_t = 4 m_c$
• speed of the car, = $v_c$
• speed of the truck, $v_t = 2v_c$

The kinetic energy of the car is calculated as;

$K_c = \frac{1}{2} m_cv_c^2$

The kinetic energy of the truck is calculated as;

$K_t = \frac{1}{2} m_tv_t^2\\\\K_t = \frac{1}{2} (4m_c)(2v_c)^2\\\\K_t = (4 \times 4)(\frac{1}{2} m_cv_c^2)\\\\K_t = 16(K_c)$

Thus, we can conclude that the relationship between the kinetic energy of truck and the car is $K_t = 16K_c$.

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