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Rama09 [41]
4 years ago
13

If a 0.50-kg block initially at rest on a frictionless, horizontal surface is acted upon by a force of 3.7 N for a distance of 6

.8 m, then what would be the block's velocity? m/s
Physics
1 answer:
mel-nik [20]4 years ago
5 0

The block's velocity is determined as 10.03 m/s.

<u>Explanation:</u>

According to work energy theorem, the work done on an object is equal to the change in kinetic energy of the object.

So, work done = Kinetic energy

\text {Force } \times \text { displacement }=\frac{1}{2} \times m \times v^{2}

Thus, the velocity can be determined as

\text {velocity}=\sqrt{\frac{2 \times \text {Force} \times \text {displacement}}{m}}

\text {velocity}=\sqrt{\frac{2 \times 3.7 \times 6.8}{0.50}}=\sqrt{\frac{50.32}{0.50}}=\sqrt{100.64}

Velocity = 10.03 m/s.

So the block's velocity is determined as 10.03 m/s.

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Help me please to do those problems...
Stells [14]

Keeping in mind that:

- Distance is the length of the total path covered by the person, regardless of the directions of the different parts of motion

- Displacement is just the distance in a straight line between the final point and the initial point

Let's apply these concepts to solve the different parts of the problem:

1. 120 m, 80 m west

The total distance is the sum of the length of the different paths:

distance = 100 + 20 = 120 m

To find the displacement, we need to find the distance between the starting point and the ending point. Assuming the starting point as

x = 0

Michelle moved 100 m westward  and 20 m eastward, so the ending point is at

ending point = 100 - 20 = 80 m (westward)

So, the displacement is

displacement = 80 - 0 = 80 m (west)

2. 6400 m, 0

The distance is equal to the length of the track multiplied by the number of laps, so:

distance = 1600 \cdot 4 = 6400 m

The track is circular, and the runner completes exactly 4 laps: it means that at the end of the motion, the runner is at her starting point. So ending point and starting point coincide, and so the displacement

displacement = ending point - starting point = 0

3. 40 m; 10 m forward

The total distance is just the sum of the lengths of the different parts of the motion, so:

distance = 20 + 5 + 10 + 5 = 40 m

The find the displacement, we need to assign signs to every direction:

Forward --> positive along the forward-backward direction

Backard --> negative along the forward-backward direction

Right --> positive direction along the right-left direction

Left --> negative direction along the right-left direction

Along the forward-backward direction, the displacement is:

20 m forward and 10 m backward, so 20 - 10 = 10 (forward)

Along the left right direction, the displacement is:

5 m right and 5 m left, so 5 - 5 = 0

So the net displacement is 10 m (forward)

4. 16 km; 4 km south

Again, t total distance is the sum of the lengths of the different parts of the motion, so:

distance = 2 + 4 + 6 + 4 = 16 km

To find the displacement, we assign signs to every direction:

North --> positive along the north-south direction

south --> negative along the north-south direction

east --> positive direction along the east-west direction

west --> negative direction along the east-west direction

Along the north-south direction, the displacement is:

2 km north and 6 km south, so 2 - 6 = -4 km (4 km south)

Along the east-west direction, the displacement is:

4 km east and 4 km west, so 4 - 4 = 0

So the net displacement is 4 km south

5. 37.7 m; 24 m

The distance covered by the skater is the length of half circumference, so given the radius

r = 12 m

The distance is

distance= \pi r = \pi \cdot (12)=37.7 m

The displacement is the distance in a straight line between the ending position and the starting position. Since the skater ends her motion halfway around the circle, the distance between the initial and final point is equal to the diameter of the circle (two times the radius). So,

displacement = 2r = 2 \cdot 12 = 24 m

3 0
3 years ago
11. A box with a mass of 1.0 kg is resting on a horizontal surface and the coefficient of friction between the block and the sur
AlladinOne [14]

Answer:

We know that the force pulling the box in the positive x direction has a magnitude of m g sin 30 . Using Newtons Second Law, F = ma , we just need to solve for a :

ma=mgsin30

a=gsin30

=(10m/s2)(0.500)

=5m/s2

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Answer:

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Explanation:

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