Gradually slow down to a safe speed well before entering the curve.
Answer:
The required potential difference is 
.
Explanation:
We know, electric field is nothing but the negative gradiant of potential. Mathematically, in three-dimension,
In one dimension, the magnitude of the electric field is
where '
' is the applied voltage and '
'is the distance through which the voltage is applied.
Given, 
.
So the required potential difference is
Vertical:
(20 m/s) sin(25º) ≈ 8.45 m/s
Horizontal:
(20 m/s) cos(25º) ≈ 18.1 m/s
Answer:
93.4 kg
Explanation:
Draw a free body diagram. There are three four forces:
Weight force mg pulling down,
Normal force N pushing up,
Friction force Nμ pushing left,
Applied force F pulling up and to the right, 30.0° above the horizontal.
Sum of forces in the y direction:
∑F = ma
N + F sin 30.0° − mg = 0
N = mg − ½ F
Sum of forces in the x direction:
∑F = ma
F cos 30.0° − Nμ = 0
½√3 F = Nμ
Substitute:
½√3 F = (mg − ½ F) μ
½√3 F / μ = mg − ½ F
½√3 F / μ + ½ F = mg
½F (√3 / μ + 1) = mg
m = F (√3 / μ + 1) / (2g)
Plug in values:
m = 410 N (√3 / 0.500 + 1) / (2 × 9.8 m/s²)
m = 93.4 kg
