A mass m = 17 kg is pulled along a horizontal floor with NO friction for a distance d =5.4 m. Then the mass is pulled up an incline that makes an angle θ = 32° with the horizontal and has a coefficient of kinetic friction μk = 0.36. The entire time the massless rope used to pull the block is pulled parallel to the incline at an angle of θ = 32° (thus on the incline it is parallel to the surface) and has a tension T =75 N.
4) What is the work done by gravity after the block has traveled a distance x = 3 m up the incline? (Where x is measured along the incline.
this what i tried but it's wrong
(17)(9.8)(sin32)(3) = 275.6
Please explain it with the correct answer
is it a significant figure issue?
all the data is 2 sig fig, while your answer is 4
To find the work done by gravity after the block has traveled a distance x = 3 m up the incline, we can use the concept of work done by a force. The work done by a force is given by the formula:
Work = Force * Distance * cos(angle)
In this case, the force is the force of gravity acting on the mass, the distance is x = 3 m, and the angle is the angle of the incline (θ = 32°). However, we need to consider that the force of gravity acts vertically downwards while the distance is measured along the incline. So, we need to resolve the force of gravity into two components: one parallel to the incline and one perpendicular to the incline.
The perpendicular component of the force of gravity is given by:
F_perpendicular = m * g * cos(θ)
where m is the mass (17 kg), g is the acceleration due to gravity (9.8 m/s^2), and θ is the angle of the incline (32°). Plugging in the values, we get:
F_perpendicular = 17 * 9.8 * cos(32°) ≈ 140.28 N
The parallel component of the force of gravity is given by:
F_parallel = m * g * sin(θ)
Plugging in the values, we get:
F_parallel = 17 * 9.8 * sin(32°) ≈ 96.13 N
Now, we can calculate the work done by gravity using the resolved force and the distance x = 3 m:
Work = F_parallel * x * cos(0°)
Since the force and the distance are parallel, the angle is 0°. Plugging in the values, we get:
Work = 96.13 N * 3 m * cos(0°) = 96.13 N * 3 m = 288.39 J
Therefore, the work done by gravity after the block has traveled a distance x = 3 m up the incline is approximately 288.39 Joules.