A block of mass 2.10 kg is pushed 2.33 m along a frictionless horizontal table by a constant 19.4 N force directed 27.5o below the horizontal. Determine the magnitude of the normal force exerted by the table.
A small block of mass m slides along the frictionless
loop-the-loop shown.
(a) If it starts from rest at P, what is the normal force
acting on it at Q?
(b) At what height above the bottom of the loop
should the block be released so that the force it
exerts against the track at the top of the loop is
equal to its weight?
To determine the magnitude of the normal force exerted by the table, you need to analyze the forces acting on the block.
1. Identify the knowns:
- Mass of the block (m): 2.10 kg
- Applied force (F): 19.4 N
- Angle of the applied force below the horizontal (θ): 27.5°
- Distance the block is pushed (d): 2.33 m
2. Resolve the applied force into horizontal and vertical components:
- F_horizontal = F * cos(θ)
- F_vertical = F * sin(θ)
- F_horizontal = 19.4 N * cos(27.5°) ≈ 16.86 N
- F_vertical = 19.4 N * sin(27.5°) ≈ 8.95 N
3. Determine the acceleration of the block:
- Since the table is frictionless, the net force acting on the block is equal to the applied force.
- Net force (F_net) = F_horizontal = 16.86 N
- F_net = m * a, where a is the acceleration of the block.
- 16.86 N = 2.10 kg * a
- a ≈ 8.029 m/s²
4. Determine the normal force exerted by the table:
- The normal force (N) balances the weight of the block and equals the mass of the block times the acceleration due to gravity (g).
- N = m * g
- N = 2.10 kg * 9.8 m/s² ≈ 20.58 N
Therefore, the magnitude of the normal force exerted by the table is approximately 20.58 N.