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3 years ago

When 142 J of work is done in pushing a box horizontally 13.3m how much force is applied

Physics

1 answer:

vampirchik [111]3 years ago

7 0

The force applied is 10.68 N

Explanation:

The work done by a force on an object is given by:

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and of the displacement

In this problem we know:

W = 142 J

d = 13.3 m

$\theta=0^{\circ}$ since the force is applied horizontally, in the direction of motion of the box

So, solving for F, we find the magnitude of the force:

$F=\frac{W}{d cos \theta}=\frac{142}{(13.3)(cos 0)}=10.68 N$

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2 years ago

A thermometer initially reading 212F is placed in a room where the temperature is 70F. After 2 minutes the thermometer reads 125

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Answer:

91.3°F

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Let T be the temperature of the thermometer at any time

T∞ be the temperature of the room = 70°F

T₀ be the initial temperature of the thermometer = 212°F

And m, c, h are all constants from the cooling law relation

From Newton's law of cooling

Rate of Heat loss by the cake = Rate of Heat gain by the environment

- mc (d/dt)(T - T∞) = h (T - T∞)

(d/dt) (T - T∞) = dT/dt (Because T∞ is a constant)

dT/dt = (-h/mc) (T - T∞)

Let (h/mc) be k

dT/(T - T∞) = -kdt

Integrating the left hand side from T₀ to T and the right hand side from 0 to t

In [(T - T∞)/(T₀ - T∞)] = -kt

(T - T∞)/(T₀ - T∞) = e⁻ᵏᵗ

(T - T∞) = (T₀ - T∞)e⁻ᵏᵗ

Inserting the known variables

(T - 70) = (212 - 70)e⁻ᵏᵗ

(T - 70) = 142 e⁻ᵏᵗ

At t = 2 minute, T = 125°F

125 - 70 = 142 e⁻ᵏᵗ

55/142 = e⁻ᵏᵗ

- kt = In (55/142) = In (0.3873)

- k(2) = - 0.9485

k = 0.4742 /min

At time t = 4 mins

kt = 0.4742 × 4 = 1.897

(T - 70) = 142 e⁻ᵏᵗ

e^(-1.897) = 0.15

T - 70 = 142 × 0.15 = 21.3

T = 91.3°F

7 0

3 years ago

When 142 J of work is done in pushing a box horizontally 13.3m how much force is applied​

When 142 J of work is done in pushing a box horizontally 13.3m how much force is applied