Solved problems in Newton’s laws of motion – Mass, and weight
1. The weight of a 1 kg mass at the surface of the Earth is… g = 9.8 m/s2
Known :
Mass (m) = 1 kg
The acceleration due to gravity at the surface of the Earth (g) = 9.8 m/s2
Wanted: weight (w)
Solution :
w = m g
m = mass (The SI unit of mass is the kilogram, kg)
g = acceleration due to gravity (The SI unit of g is m/s2)
w = weight (The SI unit of w is kg m/s2 or Newton)
Weight :
w = (1 kg)(9.8 m/s2) = 9.8 kg m/s2 = 9.8 Newton
2.
(a) Draw the force of gravity (weight) that act on the object when the object is at rest on a table, as shown in figure (a).
(b) Draw the force of gravity (weight) and it’s components that act on an object sliding down an inclined plane, as shown in figure (b)
Solution
The direction of the weight is downward toward the center of the Earth.
wx = the horizontal component of the weight and wy = the vertical component of the weight
3. The mass of a box is 1 kg and acceleration due to gravity is 9.8 m/s2. Find (a) weight (b) the horizontal component and the vertical component of the weight.
Solution
Weight : w = m g = (1 kg)(9.8 m/s2) = 9.8 kg m/s2 = 9.8 Newton
The horizontal component of the weight :
wx = w sin 30o = (9,8 N)(0,5) = 4.9 Newton
The vertical component of the weight :
wy = w cos 30o = (9.8 N)(0.5√3) = 4.9√3 Newton
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- Mass and weight
- Normal force
- Newton’s second law of motion
- Friction force
- Motion on the horizontal surface without friction force
- The motion of two bodies with the same acceleration on the rough horizontal surface with the friction force
- Motion on the inclined plane without friction force
- Motion on the rough inclined plane with the friction force
- Motion in an elevator
- The motion of bodies connected by cord and pulley
- Two bodies with the same magnitude of accelerations
- Rounding a flat curve – dynamics of circular motion
- Rounding a banked curve – dynamics of circular motion
- Uniform motion in a horizontal circle
- Centripetal force in uniform circular motion