physics help?! Find the box’s speed vf at 2.0s after you first started pushing on it.?
A box with a mass of 3.1 kg slides down a rough vertical wall. The gravitational force acting on the box is 30 N. When the box reaches a speed of 2.5 m/s, you begin pushing on one edge of the box at a 45° angle (please use degrees in your calculations throughout this problem) with a constant force of magnitude Fp = 23 N, as shown in Figure 1. At this point, there is a frictional force between the box and the wall with a magnitude of 13 N.
How fast is the box sliding 2.0 seconds after you started pushing on it? If possible, please show your work and specify the equations used in your calculations. Thank you!
1 Answers
No figure 1! I’ll assume the applied force is at 45 degrees upwards.??
Original speed 2.5m/sec^2.
UP component of applied force = (sin 45 x 23) = 16.263N.
You then produce 13N. frictional force.
Total force acting upwards = (16.263 + 13) = 29.263N.
The weight of the box is (3.1 x g) = 30.38N.
Net accelerating force now operating = (30.38 – 29.263) = 1.117N.
The acceleration is still positive.
Acceleration = (f/m) = 1.117/3.1, = 0.36m/sec^2.
After 2 secs., the box has gained (at) = 0.36 x 2) = 0.72m/sec.
Final speed after 2 secs = (2.5 + 0.72) = 3.22m/sec.
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