a mass of 5 kg descending vertically,draws up a mass of 3 kg by means of a light string passing over a pully.at the end of 4 s,the string.how much higher the 3 kg mass would go?
I think the answer is 4.9 m
Please don't cheat and use this website do your work and learn! - Troller
To answer this question, we can use the principle of conservation of mechanical energy. In this scenario, the total mechanical energy of the system (sum of potential and kinetic energies) remains constant.
Let's start by finding the initial gravitational potential energy of the system. The gravitational potential energy (PE) of an object near the Earth's surface is given by the formula:
PE = m * g * h
Where:
m is the mass (in kg),
g is the acceleration due to gravity (approximately 9.8 m/s^2),
and h is the height (in meters).
For the 5 kg mass, the initial potential energy is:
PE1 = 5 kg * 9.8 m/s^2 * 0 m (since it is at the reference height, 0 m)
PE1 = 0 J
For the 3 kg mass, we need to determine the final height it reaches after 4 seconds. To do this, we can find the acceleration of the 5 kg mass using Newton's second law:
F = m * a
The force acting on the system is the difference in the weights of the two masses:
Force = (5 kg * 9.8 m/s^2) - (3 kg * 9.8 m/s^2)
Force = 49 N - 29.4 N
Force = 19.6 N
Using this force, we can find the acceleration of the 5 kg mass:
19.6 N = 5 kg * a
a = 3.92 m/s^2
Now, let's calculate the distance traveled by the 5 kg mass in 4 seconds. We can use the equation of motion:
d = v0 * t + (1/2) * a * t^2
Where:
d is the distance traveled (in meters),
v0 is the initial velocity (in m/s),
a is the acceleration (in m/s^2),
and t is the time (in seconds).
Since the 5 kg mass starts from rest, the initial velocity is 0:
d = 0.5 * 3.92 m/s^2 * (4 s)^2
d = 0.5 * 3.92 m/s^2 * 16 s^2
d = 31.36 m
This means the 5 kg mass descends by 31.36 meters in 4 seconds. Since the string doesn't stretch, the 3 kg mass is pulled up the same distance. So, the 3 kg mass will go 31.36 meters higher.