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tatuchka [14]
2 years ago
12

1. A spring extends by 10 cm when a mass of 100 g is attached to it. What is the spring constant? (Calculate your answer in N/m)

10N/m
2. What will be the extension of this spring if the load is a) 4N and b) 75 g?
Physics
1 answer:
cupoosta [38]2 years ago
5 0

Answer:

1) k = 10 [N/m]

2) a-) x = 0.4 [m]

b)  x = 0.075 [m]

Explanation:

To be able to solve this type of problems that include springs we must use Hooke's law, which relates the force to the deformed length of the spring and in the same way to the spring coefficient.

F = k*x

where:

F = force [N] (units of Newtons]

k = spring constant  [N/m]

x = distance = 10 [cm] = 0.1 [m]

Now, the weight is equal to the product of the mass by the gravity

W = m*g = F

where:

m = mass = 100 [g] = 0.1 [kg]

g = gravity acceleration = 10 [m/s²]

F = 0.1*10 = 1 [N]

Now clearing k

k = F/x

k = 1/0.1

k = 10 [N/m]

2)

a ) if the force is 4 [N]

clearing x

x = F/k

x = 4/10

x = 0.4 [m]

m = 75 [g] = 0.075 [kg]

W = m*g = F

F = 0.075*10 = 0.75 [N]

x = .75/10

x = 0.075 [m]

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A horizontal force of 12N is applied to 1.5kg of block which rests on a horizontal surface. If the coefficient friction is 0.3,
Rama09 [41]

Answer:

7 m/s²

Explanation:

From the question given above, the following data were:

Force applied (Fₐ) = 15 N

Mass (m) of block = 1.5 Kg

Acceleration due to gravity (g) = 10 m/s²

Coefficient of friction (μ) = 0.3

Acceleration (a) of block =?

Next, we shall determine the frictional force. This can be obtained as follow:

Mass (m) of block = 1.5 Kg

Acceleration due to gravity (g) = 10 m/s²

Coefficient of friction (μ) = 0.3

Normal reaction (R) = mg = 1.5 × 10 = 15 N

Frictional force (Fբ) =?

Fբ = μR

Fբ = 0.3 × 15

Fբ = 4.5 N

Next, we shall determine the net force acting on the block. This can be obtained as follow:

Force applied (Fₐ) = 15 N

Frictional force (Fբ) = 4.5 N

Net force (Fₙ) =.?

Fₙ = Fₐ – Fբ

Fₙ = 15 – 4.5

Fₙ = 10.5 N

Finally, we shall determine the acceleration produced in the block. This can be obtained as follow:

Mass (m) of block = 1.5 Kg

Net force (Fₙ) = 10.5 N

Acceleration (a) of block =?

Fₙ = ma

10.5 = 1.5 × a

Divide both side by a

a = 10.5 / 1.5

a = 7 m/s²

Therefore, the acceleration produced in the block is 7 m/s²

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