To solve the problem step-by-step, we will use the principles of momentum and the concept of thrust force due to the sand being dropped into the cart.
### Step 1: Identify the given values
- Mass of the cart, \( m_0 = 150 \, \text{kg} \)
- Force applied, \( F = 10 \, \text{N} \)
- Rate of sand being dropped, \( \mu = 100 \, \text{g/s} = 0.1 \, \text{kg/s} \)
- Mass of sand when the cart has 100 kg of sand, \( m = 100 \, \text{kg} \)
### Step 2: Determine the total mass of the system
The total mass of the system when the cart has 100 kg of sand is:
\[
M = m_0 + m = 150 \, \text{kg} + 100 \, \text{kg} = 250 \, \text{kg}
\]
### Step 3: Set up the equation of motion
The thrust force due to the sand being dropped is given by:
\[
\text{Thrust force} = \mu v
\]
The net force acting on the system can be expressed as:
\[
F - \mu v = M \frac{dv}{dt}
\]
Where \( dv/dt \) is the acceleration of the system.
### Step 4: Rearranging the equation
Rearranging gives:
\[
F - \mu v = \frac{d(Mv)}{dt}
\]
Since the mass is changing due to the sand being added, we can express this as:
\[
F - \mu v = \frac{d}{dt}(M v) = M \frac{dv}{dt} + v \frac{dm}{dt}
\]
Here, \( \frac{dm}{dt} = \mu \).
### Step 5: Substitute values
Substituting the values we have:
\[
10 - 0.1v = 250 \frac{dv}{dt} + v(0.1)
\]
This simplifies to:
\[
10 - 0.1v = 250 \frac{dv}{dt} + 0.1v
\]
Combining terms gives:
\[
10 = 250 \frac{dv}{dt} + 0.2v
\]
### Step 6: Solve for velocity
Now we need to integrate this equation. We can express it as:
\[
\frac{dv}{dt} = \frac{10 - 0.2v}{250}
\]
Integrating both sides will give us the velocity as a function of time.
### Step 7: Find the final velocity when the mass is 250 kg
Assuming that the system starts from rest, we can integrate to find the final speed when the cart has 100 kg of sand.
Using the relationship:
\[
v = \frac{Ft}{M + \mu t}
\]
Substituting \( F = 10 \, \text{N} \), \( M = 150 \, \text{kg} + 100 \, \text{kg} = 250 \, \text{kg} \), and \( \mu = 0.1 \, \text{kg/s} \):
\[
v = \frac{10t}{150 + 0.1t}
\]
To find the time when the cart has 100 kg of sand:
\[
100 = 0.1t \implies t = 1000 \, \text{s}
\]
Now substituting \( t = 1000 \, \text{s} \) into the velocity equation:
\[
v = \frac{10 \times 1000}{150 + 0.1 \times 1000} = \frac{10000}{150 + 100} = \frac{10000}{250} = 40 \, \text{m/s}
\]
### Final Answer
Thus, the speed of the system when the cart has 100 kg of sand in it is:
\[
\boxed{40 \, \text{m/s}}
\]
