1.- Al aplicar una fuerza de 60 Newtons sobre un cuerpo, se produce una aceleración de 4 m/s2 calcula el valor de su masa.
1 answer:
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Answer:
0.56
Explanation:
Let the coefficient of friction is μ.
m = 4.3 kg, θ = 30 degree, initial velocity, u = 0, s = 2.7 m, t = 5.8 s
By the free body diagram,
Normal reaction, N = mg Cosθ = 4.3 x 9.8 x Cos 30 = 36.49 N
Friction force, f = μ N = 36.49 μ
Net force acting on the block,
Fnet = mg Sinθ - f = 4.3 x 9.8 x Sin 30 - 36.49 μ
Fnet = 21.07 - 36.49μ
Net acceleartion, a = Fnet / m
a = (21.07 - 36.49μ) / 4.3
use second equation of motion
s = ut + 1/2 a t^2
2.7 = 0 + 1/2 x (21.07 - 36.49μ) x 5.8 x 5.8 / 4.3
By solving we get
μ = 0.56
Answer:
= 33.33 cm
Explanation:
Given:
When mass, =21 kg
distance travelled is = 140 cm
When mass, =5 kg
distance travelled is = ?
Hooke's law state that within elastic limit, when an external force is applied to a body, the body gets deformed and when the force is released the gets back to its original form.
Therefore according to the question,
= 33.33 cm
Distance travelled is 33.33 cm when mass is 5 kg.