Answer:
303.29N and 1.44m/s^2
Explanation:
Make sure to label each vector with none, mg, fk, a, FN or T
Given
Mass m = 68.0 kg
Angle θ = 15.0°
g = 9.8m/s^2
Coefficient of static friction μs = 0.50
Coefficient of kinetic friction μk =0.35
Solution
Vertically
N = mg - Fsinθ
Horizontally
Fs = F cos θ
μsN = Fcos θ
μs( mg- Fsinθ) = Fcos θ
μsmg - μsFsinθ = Fcos θ
μsmg = Fcos θ + μsFsinθ
F = μsmg/ cos θ + μs sinθ
F = 0.5×68×9.8/cos 15×0.5×sin15
F = 332.2/0.9659+0.5×0.2588
F =332.2/1.0953
F = 303.29N
Fnet = F - Fk
ma = F - μkN
a = F - μk( mg - Fsinθ)
a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0
303.29-0.35( 666.4 - 303.29*0.2588)/68.0
303.29-0.35(666.4-78.491)/68.0
303.29-0.35(587.90)/68.0
(303.29-205.45)/68.0
97.83/68.0
a = 1.438m/s^2
a = 1.44m/s^2
Answer:
55N
Explanation:
Using Newton's second law of motion:
F=ma
Force=mass × acceleration
F=25×2.2
F=55N
So 55 Newtons are needed
The acceleration due to gravity near the surface of the planet is 27.38 m/s².
<h3>
Acceleration due to gravity near the surface of the planet</h3>
g = GM/R²
where;
- G is universal gravitation constant
- M is mass of the planet
- R is radius of the planet
- g is acceleration due to gravity = ?
g = (6.626 x 10⁻¹¹ x 2.81 x 5.97 x 10²⁴) / (6371 x 10³)²
g = 27.38 m/s²
Thus, the acceleration due to gravity near the surface of the planet is 27.38 m/s².
Learn more about acceleration due to gravity here: brainly.com/question/88039
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