To solve the problem step by step, we will analyze the motion of the cart and how the mass changes over time due to the sand spilling out.
### Step 1: Understand the initial conditions
- The initial mass of the cart with sand, \( m_0 = 1800 \, \text{kg} \).
- The force acting on the cart, \( F = 120 \, \text{N} \).
- The rate at which sand spills out, \( \frac{dm}{dt} = -0.5 \, \text{kg/s} \) (negative because mass is decreasing).
- The time duration for which the cart moves, \( t = 20 \, \text{min} = 1200 \, \text{s} \).
### Step 2: Determine the mass of the cart after 20 minutes
The mass of the cart at any time \( t \) can be expressed as:
\[
m(t) = m_0 - \left(\frac{dm}{dt} \cdot t\right) = 1800 - (0.5 \cdot t)
\]
After \( t = 1200 \, \text{s} \):
\[
m(1200) = 1800 - (0.5 \cdot 1200) = 1800 - 600 = 1200 \, \text{kg}
\]
### Step 3: Relate force, mass, and acceleration
According to Newton's second law:
\[
F = m(t) \cdot a(t)
\]
Where \( a(t) = \frac{dv}{dt} \). Therefore:
\[
a(t) = \frac{F}{m(t)} = \frac{120}{m(t)}
\]
Substituting for \( m(t) \):
\[
a(t) = \frac{120}{1800 - 0.5t}
\]
### Step 4: Set up the differential equation
We have:
\[
\frac{dv}{dt} = \frac{120}{1800 - 0.5t}
\]
### Step 5: Integrate to find velocity
We need to integrate both sides:
\[
\int dv = \int \frac{120}{1800 - 0.5t} dt
\]
The left side integrates to \( v \), and for the right side, we can use substitution:
Let \( u = 1800 - 0.5t \), then \( du = -0.5 dt \) or \( dt = -2 du \).
Changing the limits:
- When \( t = 0 \), \( u = 1800 \).
- When \( t = 1200 \), \( u = 1200 \).
Thus:
\[
\int \frac{120}{u} (-2) du = -240 \ln|u| \Big|_{1800}^{1200}
\]
Calculating this gives:
\[
v = -240 \left( \ln(1200) - \ln(1800) \right) = 240 \ln\left(\frac{1800}{1200}\right) = 240 \ln\left(\frac{3}{2}\right)
\]
### Step 6: Final velocity
The final velocity after 20 minutes is:
\[
v = 240 \ln\left(\frac{3}{2}\right)
\]
