1. It takes 190 kJ of work to accelerate a car from 21.0 m/s to 29.0 m/s. What is the car's mass?
2. A 32-kg girl is bouncing on a trampoline. During a certain interval after she leaves the surface of the trampoline, her kinetic energy decreases to 205 J from 435 J. How high does she rise during this interval? Neglect air resistance.
1. 190,000 = (M/2)(29^2 - 21^2)
Solve for M in kg
2. K.E. loss = P.E. gain
(425 - 235) = M g H
Solve for H, the distance she rises during that interval.
(She will rise still farther until her kinetic energy is zero.)
To solve both questions, we need to use the principle of work-energy theorem, which states that the net work done on an object is equal to its change in kinetic energy.
1. To find the car's mass, we can use the equation for work:
Work = (1/2) * mass * (final velocity^2 - initial velocity^2)
Given:
Work = 190 kJ = 190,000 J
Initial velocity (u) = 21.0 m/s
Final velocity (v) = 29.0 m/s
Substituting these values into the equation, we can solve for mass:
190,000 = (1/2) * mass * (29.0^2 - 21.0^2)
190,000 = (1/2) * mass * (841 - 441)
190,000 = (1/2) * mass * 400
mass = (2 * 190,000) / 400
mass = 950 kg
Therefore, the car's mass is 950 kg.
2. To find the height she rises during the interval, we can use the equation for gravitational potential energy:
Potential Energy (PE) = mass * gravity * height
Given:
Mass (m) = 32 kg
Initial kinetic energy (KE_initial) = 435 J
Final kinetic energy (KE_final) = 205 J
Gravity (g) = 9.8 m/s^2
Since the girl's initial and final velocities are not given, we'll use the difference in kinetic energy to determine her change in height.
Change in kinetic energy (ΔKE) = KE_final - KE_initial
ΔKE = 205 J - 435 J
ΔKE = -230 J (negative since kinetic energy is decreasing)
The change in kinetic energy is equal to the negative of the change in potential energy.
ΔPE = -ΔKE
ΔPE = m * g * Δh
-230 J = 32 kg * 9.8 m/s^2 * Δh
-230 J = 313.6 kg·m^2/s^2 * Δh
Solving for Δh:
Δh = -230 J / (313.6 kg·m^2/s^2)
Δh = -0.733 m
Since height cannot be negative, the girl rises by 0.733 meters during this interval.
Therefore, the girl rises by 0.733 meters.