At t=1.0s, a 0.60kg object is falling with a speed of 6.0 m/s. At t=2.0s, it has a kinetic energy of 25 J.
a) What is the kinetic energy of the object at t=1.0s?
b)What is the speed of the object at t=2.0s?
c) How much work was done on the object between t=1.0s and t=2.0s?
a)m=0.6, v=6
k=1/2*m*v^2
b)m=0.6, k=25
k=1/2*m*v^2
v=sqrt(2*k/m)
c)w=k_2-k_1
a) Well, at t=1.0s, the kinetic energy of the object is the funniest thing ever... zero! That's right, zero kinetic energy, because it hasn't started moving yet. It's just chilling, like a couch potato.
b) Now, at t=2.0s, the object is all pumped up and has a kinetic energy of 25 J. I guess it's been hitting the gym. Maybe it's been doing some speed workouts too...
c) As for the work done on the object between t=1.0s and t=2.0s, let's see... the object is falling, so gravity is doing some work there. But, let's be honest, it's not like gravity has a job or something. I mean, who hired gravity? And seriously, what's its paycheck like? Anyway, the work done by gravity is equivalent to the change in kinetic energy. So, the work done is 25 J, just like the kinetic energy at t=2.0s. It's like the universe giving you a high-five for physics. Pretty cool, huh?
a) To find the kinetic energy at t=1.0s, we can use the formula for kinetic energy:
Kinetic energy = (1/2) * mass * (speed)^2
Given:
mass = 0.60 kg
speed = 6.0 m/s
Kinetic energy at t=1.0s = (1/2) * 0.60 kg * (6.0 m/s)^2
Step 1: Square the speed
= (1/2) * 0.60 kg * 36.0 m^2/s^2
Step 2: Multiply the mass and the squared speed
= 0.50 * 36.0 kg * m^2/s^2
Step 3: Simplify the equation
= 18 J
Therefore, the kinetic energy of the object at t=1.0s is 18 J.
b) To find the speed at t=2.0s, we need to use the given kinetic energy at t=2.0s and rearrange the formula for kinetic energy:
Kinetic energy = (1/2) * mass * (speed)^2
Given:
mass = 0.60 kg
Kinetic energy at t=2.0s = 25 J
25 J = (1/2) * 0.60 kg * (speed)^2
Step 1: Divide both sides by (1/2) * mass
(speed)^2 = (25 J) / [(1/2) * 0.60 kg]
Step 2: Simplify the expression on the right
(speed)^2 = 25 J / 0.30 kg
Step 3: Calculate the value of (speed)^2
(speed)^2 = 83.33 m^2/s^2
Step 4: Take the square root of both sides to solve for speed
speed = √(83.33 m^2/s^2)
To three significant figures, the speed is approximately 9.12 m/s.
Therefore, the speed of the object at t=2.0s is approximately 9.12 m/s.
c) To find the work done on the object between t=1.0s and t=2.0s, we need to use the work-energy theorem:
Work done = Change in kinetic energy
Given:
Change in kinetic energy = 25 J - 18 J = 7 J
Therefore, the work done on the object between t=1.0s and t=2.0s is 7 J.
To find the answers to these questions, we need to use the formulas related to kinetic energy (KE), speed (v), and work (W).
a) The kinetic energy of an object is given by the formula KE = (1/2)mv^2, where m is the mass of the object and v is its velocity. Given that the mass is 0.60kg and the speed at t=1.0s is 6.0 m/s, we can calculate the kinetic energy using the formula:
KE = (1/2)(0.60 kg)(6.0 m/s)^2
Simplifying the equation, we get:
KE = (1/2)(0.60 kg)(36 m^2/s^2)
KE = 10.8 J
Therefore, the kinetic energy of the object at t=1.0s is 10.8 J.
b) To find the speed of the object at t=2.0s, we can use the formula for kinetic energy:
KE = (1/2)mv^2
Given that the kinetic energy at t=2.0s is 25 J and the mass remains the same (0.60kg), we can rearrange the formula:
25 J = (1/2)(0.60 kg)(v^2)
Simplifying the equation, we get:
v^2 = 83.33 m^2/s^2
Taking the square root of both sides, we find:
v ≈ 9.13 m/s
Therefore, the speed of the object at t=2.0s is approximately 9.13 m/s.
c) The work done on an object is given by the formula W = ΔKE, where ΔKE is the change in kinetic energy. To calculate the work done between t=1.0s and t=2.0s, we subtract the initial kinetic energy at t=1.0s from the final kinetic energy at t=2.0s:
W = KE_final - KE_initial
Given that the initial kinetic energy is 10.8 J (from part a) and the final kinetic energy is 25 J, we can substitute these values into the equation:
W = 25 J - 10.8 J
W ≈ 14.2 J
Therefore, the work done on the object between t=1.0s and t=2.0s is approximately 14.2 J.