4. a cinder block is sitting on a platform 200 m high. it weighs 17.9 kg. the block has
__________ energy.
calculate it.

5) angela uses a force of 200 newtons to lift her grocery bag while doing 50 joules of work.
how far did she lift the grocery bags?

6) the baseball player does 1500 joules of work when hitting a baseball into left field.
assuming the baseball landed 120 meters away from home plate, how much force did the
player use to hit the ball?

7) sketch a graph that shows the amount of work done on a cart of the same mass at different
angles of a ramp that reach the same height.

8) sketch a graph that shows the amount of force done on a cart of the same mass at different
angles of a ramp that reach the same height.

Question

4. a cinder block is sitting on a platform 200 m high. it weighs 17.9 kg. the block has
__________ energy.
calculate it.

5) angela uses a force of 200 newtons to lift her grocery bag while doing 50 joules of work.
how far did she lift the grocery bags?

6) the baseball player does 1500 joules of work when hitting a baseball into left field.
assuming the baseball landed 120 meters away from home plate, how much force did the
player use to hit the ball?

7) sketch a graph that shows the amount of work done on a cart of the same mass at different
angles of a ramp that reach the same height.

8) sketch a graph that shows the amount of force done on a cart of the same mass at different
angles of a ramp that reach the same height.

Accepted Answer

### Question 4
# Explanation:
## Step1: Identify Energy Type
The cinder block is at a height, so it has gravitational potential energy. The formula for gravitational potential energy is $PE = mgh$, where $m$ is mass, $g$ is acceleration due to gravity ($9.8\ m/s^2$), and $h$ is height.
## Step2: Plug in Values
Given $m = 17.9\ kg$, $g = 9.8\ m/s^2$, $h = 200\ m$. Calculate $PE = 17.9	imes9.8	imes200$.
First, $17.9	imes9.8 = 175.42$. Then, $175.42	imes200 = 35084$.
# Answer:
The block has gravitational potential energy, and its value is $35084\ J$ (or $3.5084	imes10^4\ J$).

### Question 5
# Explanation:
## Step1: Recall Work Formula
The formula for work is $W = Fd$, where $W$ is work, $F$ is force, and $d$ is distance. We need to find $d$, so rearrange the formula to $d=\frac{W}{F}$.
## Step2: Substitute Values
Given $W = 50\ J$, $F = 200\ N$. So $d=\frac{50}{200}=0.25\ m$.
# Answer:
She lifted the grocery bag $0.25\ meters$.

### Question 6
# Explanation:
## Step1: Recall Work Formula
Using $W = Fd$, we need to find $F$, so rearrange to $F=\frac{W}{d}$.
## Step2: Substitute Values
Given $W = 1500\ J$, $d = 120\ m$. So $F=\frac{1500}{120}=12.5\ N$.
# Answer:
The player used a force of $12.5\ Newtons$.

### Question 7 (Graph Sketch Explanation)
# Explanation:
## Step1: Understand Work on Ramp
The work done to lift a cart to height $h$ is $W = mgh$ (since work against gravity depends on height, not ramp angle for same height). But when using a ramp, the force is $F = mg\sin	heta$ (where $	heta$ is ramp angle), and distance along ramp is $d=\frac{h}{\sin	heta}$. So work $W = Fd=(mg\sin	heta)(\frac{h}{\sin	heta})=mgh$, which is constant for same mass and height.
## Step2: Graph Description
The x - axis is the ramp angle ($	heta$), y - axis is work done. The graph will be a horizontal line (since work $W = mgh$ is constant for different $	heta$ as long as $m$, $g$, $h$ are constant).
# Answer:
The graph is a horizontal line (y - value constant) with x - axis as ramp angle and y - axis as work done.

### Question 8 (Graph Sketch Explanation)
# Explanation:
## Step1: Force on Ramp
The force required to move the cart up the ramp at constant speed (neglecting friction) is $F = mg\sin	heta$, where $	heta$ is the ramp angle, $m$ is mass, $g$ is gravity. As $	heta$ increases from $0^{\circ}$ to $90^{\circ}$, $\sin	heta$ increases from 0 to 1.
## Step2: Graph Description
The x - axis is the ramp angle ($	heta$), y - axis is force. The graph will be a curve (or line, since $F$ is proportional to $\sin	heta$) that starts at 0 when $	heta = 0^{\circ}$ and increases to $mg$ when $	heta = 90^{\circ}$.
# Answer:
The graph has x - axis as ramp angle, y - axis as force. It is an increasing curve (or line) from 0 (at $	heta = 0^{\circ}$) to $mg$ (at $	heta = 90^{\circ}$), following $F = mg\sin	heta$.

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Question 4

Step1: Identify Energy Type

Step2: Plug in Values

Step1: Recall Work Formula

Step2: Substitute Values

Step1: Recall Work Formula

Step2: Substitute Values

Answer:

Question 5