pete and john play a game of tug f war on a frictionless icy curface. pete weighs 539 N and john weights 392 N. during the course of the game, john accelerates toward pete at a rate of 3 m/s^2
a)what's the magnitude of the force that peter exerts on john?
b)what's the magnitude of pete's acceleration toward john?
I am not sure what formula to use, pleaase help!
PHYSICS - drwls, Saturday, January 23, 2010 at 5:34pm
The center of mass will remain in the same place in this frictionless situation
a) Use F = ma
Solve for the F
b) Also use F = ma on Pete, with the same F, and solve for a(of Pete).
The F will be the same on both, but in opposite directions.
QUESTION: DO I HAVE TO CONVERT THE WEIGHT TO MASS BY DIVIDING THE N BY 9.8M/S^2? THANKYOU.
Yes, you should convert weight to mass.
Yes, you need to convert the weight to mass by dividing the force (in Newtons) by the acceleration due to gravity (approximately 9.8 m/s^2) in order to use the formulas correctly.
Let's start with the first question:
a) To find the magnitude of the force that Pete exerts on John, you can use Newton's second law, which states that force (F) is equal to mass (m) multiplied by acceleration (a). In this case, John's mass (m) is 392 N / 9.8 m/s^2 ≈ 40 kg, and his acceleration (a) is given as 3 m/s^2. So the force that John exerts on Pete is F = m x a = 40 kg x 3 m/s^2 = 120 N. Since the forces are equal and opposite in a tug of war, Pete exerts the same magnitude of force, which is 120 N.
b) To find the magnitude of Pete's acceleration toward John, you can again use Newton's second law. Pete's mass (m) is 539 N / 9.8 m/s^2 ≈ 55 kg. Since the force exerted on Pete by John is 120 N, the acceleration of Pete (a) can be calculated as a = F / m = 120 N / 55 kg ≈ 2.18 m/s^2. Therefore, the magnitude of Pete's acceleration toward John is approximately 2.18 m/s^2.
Remember, in physics, it is important to convert units correctly and use the appropriate formulas to find the desired information.