1. What is the weight of a 10-kg object? 2. A 30-kg sled is being pulled along i

Question

1. What is the weight of a 10-kg object?
2. A 30-kg sled is being pulled along ice-covered ground. How much friction is there between the sled and the ice, = 0.1.
3. A 7-kg sign hangs from a single cable attached to the ceiling. How much tension is in the cable?
4. What is the weight for each of the following?
a) 2.00 kg bag of sugar
b) 110 kg football player
c) 870 kg Honda Civic
5. What is the mass of a man that weighs 726 N?
6. How much does a 74-kg man weigh
a) in London? g = 9.81 m/s2
b) in Hong Kong? g = 9.79 m/s2
c) 6,400 km above the surface of the Earth? g = 2.45 m/s2
7. How much friction is there between a 74-kg man on waxed skis and snow ( = 0.05)?
8. A falling 12-kilogram object has an acceleration of 7 m/s2. How much air resistance is acting on the object?
9. What is the acceleration of a falling 75-kilogram mass if it experiences 735 N of air resistance?
10. Look at the diagram below. Calculate the tension in each rope.
11. a) Convert you weight from pounds to newtons. pounds × 4.4
b) Calculate your mass. This will be in kilograms.
c) Calculate your weight if you were on the Moon. g = 1.63 m/s2
d) Convert your Moon weight from newtons to pounds. newtons ÷ 4.4

Explanation / Answer

1) Weight is the effect of gravity on an object. Gravity = 9.8 m/s^2 Therefore, the weight of a 10kg object is (10)(9.8) = 98 kg*m/s^2 or 98 Newtons (N) 2) The coefficient of friction = .1. Friction = the normal force * coefficient. Fricion Force = (.1)(30kg)(9.8 m/s^2) = 29.4 N 3) If you draw a Free Body Diagram, The forces acting down are (7kg)(9.8m/s^2). To keep this in equilibrium, the forces upwards will be the same, which is the tension in the rope. Tension = (7)(9.8)= 68.6 N 4) a) Weight = (2kg)(9.8 m/s^2) = 19.6 N b) Weight = (110kg)(9.8 m/s^2) = 1078 N c) Weight = (870kg)(9.8 m/s^2) = 8526 N 5) The man weighs 726 N. W=mass * acceleration acceleration = 9.8 m/s^2 726/9.8 = 74.08 kg 6) 74 kg * 9.8 m/s^2 = 725.2 N 7) Coefficient of Friction = .05 Friction Force = Coefficient * normal force Normal force = the force the ground exerts on the man (upwards) normal force = (74kg)(9.8 m/s^2) = 725.2 N Friction Force = (725.2)(.05)= 36.26 N 8) Draw Free body Diagram The net force is acting downward which equals (12kg)(7m/s^2) = 84 N The Total force downward is (12kg)(9.8m/s^2) = 117.6 N Net Force = Downward Force - Upward Force 84N = 117.6 N - Air resistance Air resistance Force = 33.6 N 9) Draw Free body Diagram. Forces Downward are (75)(9.8) = 735 N Forces Upward are 735 N Subtract the forces to get 0 F= mass*acceleration Therefore, acceleration = 0 10) Dont have Diagram. Remember, tension is Force in the rope. Force = mass * acceleration. Drawing a Free body diagram usually helps a lot. 11) a) 1 lb = 4.4 Newtons Lets say you weigh "150" pounds (if not, just enter your weight and follow same steps) (150)* (4.4) = 660 Newtons b) Divide by 9.8 m/s^2 660/9.8 = 67.34 kg c) if your mass is 67.34 kg, then multiply by the moons gravity (67.34 kg)*(1.63 m/s^2) = 109.76 N d) (109.76 N) / (4.4) = 24.95 Pounds

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.