An object with a mass of #2 kg# is on a plane with an incline of #pi/8 #. If the object is being pushed up the plane with # 1 N # of force, what is the net force on the object?

Question

An object with a mass of #2   kg# is on a plane with an incline of #pi/8   #. If the object is being pushed up the plane with # 1  N # of force, what is the net force on the object?

Charles Aeschliman · Accepted Answer

#F_n=6,508N# #alpha=pi/8*180/pi=22,5^o# #F=m*g*sin alpha# #F=2*9,81*0,382683432365# #F=7,508 N# #F_n=7,508-1# #F_n=6,508N#

Adam Arnold · Answer

undefined The net force on the object can be calculated using the equation: 
$$F_{net} = F_{applied} - F_{gravity \ parallel \ incline} = F_{applied} - m \cdot g \cdot sin(\theta)$$

Given:
- $m = 2 \, \text{kg}$
- $F_{applied} = 1 \, \text{N}$
- $\theta = \frac{\pi}{8}$
- $g$ is the acceleration due to gravity ($9.81 \, \text{m/s}^2$)

Plug in the values:
$$F_{net} = 1 \, \text{N} - (2 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \cdot sin(\frac{\pi}{8}))$$

$$F_{net} = 1 \, \text{N} - (2 \cdot 9.81 \cdot sin(22.5^\circ))$$

$$F_{net} \approx 1 \, \text{N} - (2 \cdot 9.81 \cdot 0.3827)$$

$$F_{net} \approx 1 \, \text{N} - 7.53 \, \text{N}$$

$$F_{net} \approx -6.53 \, \text{N}$$

Therefore, the net force on the object is approximately $ -6.53 \, \text{N}$.