A cart is pulled by a force of [tex]$250 N$[/tex] at an angle of [tex]$35^{\circ}$[/tex] above the horizontal. The cart accelerates at [tex]$1.4 \, \text{m/s}^2$[/tex].

To the nearest whole number, what is the mass of the cart?

The free-body diagram shows the forces acting on the cart. [tex]$\square \, \text{kg}$[/tex].



Answer :

Sure! Let's find the mass of the cart by breaking down the steps:

1. Given values:
- Force applied ([tex]\( F \)[/tex]) = 250 Newtons (N)
- Angle of force ([tex]\( \theta \)[/tex]) = 35 degrees
- Acceleration ([tex]\( a \)[/tex]) = 1.4 meters per second squared ([tex]\( m/s^2 \)[/tex])

2. Resolve the horizontal component of the force:
The force has a vertical and a horizontal component due to the angle. For this problem, we are interested in the horizontal component of the force which contributes to the acceleration of the cart.

The horizontal component ([tex]\( F_{\text{horizontal}} \)[/tex]) of the force can be calculated using the cosine of the angle:
[tex]\[ F_{\text{horizontal}} = F \cdot \cos(\theta) \][/tex]

Substituting the given values:
[tex]\[ F_{\text{horizontal}} = 250 \cdot \cos(35^{\circ}) \approx 204.788 \][/tex]

3. Use Newton's Second Law to find the mass:
Newton's Second Law states:
[tex]\[ F = ma \][/tex]
where [tex]\( F \)[/tex] is the net force acting on an object, [tex]\( m \)[/tex] is the mass of the object, and [tex]\( a \)[/tex] is the acceleration.

We can rearrange this formula to solve for the mass ([tex]\( m \)[/tex]):
[tex]\[ m = \frac{F_{\text{horizontal}}}{a} \][/tex]

Substituting the known values:
[tex]\[ m = \frac{204.788}{1.4} \approx 146.277 \][/tex]

4. Round to the nearest whole number:
[tex]\[ m \approx 146 \][/tex]

Thus, the mass of the cart is [tex]\( 146 \)[/tex] kg. So, to the nearest whole number, the mass of the cart is 146 kg.

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