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3 years ago

A child carries a 3N book at a constant velocity 4 meters across a horizontal floor. What is the net work done?

1 answer:

5 0

Note that 4 m (meters) has the units of distance, not velocity.

By definition,
Work = Force x Distance
Therefore if a 3 N book is carried over a distance of 4 m, the work done is
3 *3 4 = 12 J

Answer: 12 J

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What is the net force 100n to the left 100 n to the right

ANEK [815]

Answer:

I believe 0 N

Explanation:

100 to the left can be said to be -100

100 to the right can be said to be +100

+100 + -100 = 0

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2 years ago

A 1.35 kg block at rest on a tabletop is attached to a horizontal spring having constant 19.8 n/m. the spring is initially unstr

Ket [755]

Horizontal force is applied to the object causing the spring to stretch. the acceleration of gravity is 9.8 m/s

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3 years ago

In a collision, a 15 kg object moving with a velocity of 3 m/s transfers some of its momentum to a 5 kg object. What would be th

Misha Larkins [42]

The key to solve this problem is the conservation of momentum. The momentum of an object is defined as the product between the mass and the velocity, and it's usually labelled with the letter $p$ :

$p=mv$

The total momentum is the sum of the momentums. The initial situation is the following:

$m_A=15,\quad v_A=3,\quad m_B=5,\quad v_B=0$

(it's not written explicitly, but I assume that the 5-kg object is still at the beginning).

So, at the beginning, the total momentum is

$p=m_Av_A+m_Bv_B=15\cdot 3+5\cdot 0=45$

At the end, we have

$m_A=15,\quad v_A=1,\quad m_B=5,\quad v_B=x$

(the mass obviously don't change, the new velocity of the 15-kg object is 1, and the velocity of the 5-kg object is unkown)

After the impact, the total momentum is

$p=m_Av_A+m_Bv_B=15\cdot 1+5\cdot x=15+5x$

Since the momentum is preserved, the initial and final momentum must be the same. Set an equation between the initial and final momentum and solve it for $x$ , and you'll have the final velocity of the 5-kg object.

4 0

3 years ago

A block slides down a frictionless plane that makes an angle of 24.0° with the horizontal. What is the

irga5000 [103]

Answer:

F = m g sin theta force accelerating block

m a = m g sin theta

a = 9.8 sin 24 = 3.99 m/sec^2

7 0

2 years ago

Two forces, F? 1 and F? 2, act at a point, as shown in the picture. (Figure 1) F? 1 has a magnitude of 9.20 N and is directed at

Stolb23 [73]

Answer:

a. Fx = -8.089 N b. Fy = 3.525 N c. 8.824 N d. 336.45°

Explanation:

Since F₁ = 9.2 N and acts at 57° above the negative axis in the second quadrant, its x-component is -F₁cos57° and its y- component is F₁sin57°

Since F₁ = 5.2 N and acts at 53.7° below the negative axis in the third quadrant, its x-component is -F₂cos53.7° and its y- component is -F₂sin53.7°

Part A

What is the x component Fx of the resultant force?

The x component of the resultant force Fx = -F₁cos57° + -F₂cos53.7° = -9.2cos57° + (-5.2cos53.7°) = (-5.011 - 3.078) N = -8.089 N

Part B

What is the y component Fy of the resultant force?

The y component Fy of the resultant force = F₁sin57° + -(F₂sin53.7°) = 9.2sin57° - 5.2sin53.7° = (7.716 - 4.191) N = 3.525 N

Part C

What is the magnitude F of the resultant force?

The magnitude F of the resultant force = √(Fx² + Fy²)

F = √(-8.089² N + 3.525² N) = √65.432 + 12.426 = √77.858 = 8.824 N

Part D

What is the angle ? that the resultant force forms with the negative x axis?

The angle the resultant force makes with the negative x axis is given by

θ = tan⁻¹(Fy/Fx) = tan⁻¹(3.525/-8.089) = tan⁻¹-0.4358 = -23.55°.

To measure it from the negative x axis, we add 360. So, our angle = 360 -23.55 = 336.45°

7 0

2 years ago