1) A 120 N force is applied at an angle of 30.0 degrees above the horizontal to

Question

1) A 120 N force is applied at an angle of 30.0 degrees above the horizontal to a 4.00 kg box. The box moves a horizontal distance of 8.00 meters. The change in the kinetic energy of the box is
A) 667 J.
B) 750 J.
C) 831 J.
D) 890 J.
E) 920 J.

2) A 3.00 kg pendulum bob on a string 2.00 m long is released with a velocity of 2.00 m/s when the support string makes an angle of 45.0 degrees with the vertical. What is the tension in the string at the highest point of its motion?
A) 12.5 N
B) 17.8 N
C) 21.4 N
D) 28.9 N
E) 31.6 N

Explanation / Answer

1.
we have,

Change in kinetic energy = net work = 120cos(30) x 8 = 831.38 Joules

Option - C is correct !!!

2.
Tension = T
T = mgcos + mv^2/R

= 3 x 9.8 cos(45) + 3 x 2^2/2

= 32.789 Newton

Option - E is correct !!

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