Answer with Explanation:
We are given that
A.Mass,m=12 kg
[tex]\theta=53^{\circ}[/tex]
[tex]\mu_k=0.16[/tex]
Speed,v=1.5m/s
Net force in x direction must be zero
[tex]F_{net}=0[/tex]
[tex]Fsin\theta-f=0[/tex]
[tex]Fsin\theta=f[/tex]
Net force in y direction
[tex]N-mg-Fcos\theta=0[/tex]
[tex]N=mg+Fcos\theta[/tex]
[tex]f=\mu_kN=\mu_k(mg+Fcos\theta)[/tex]
[tex]Fsin\theta=\mu_k(mg+Fcos\theta)[/tex]
[tex]Fsin\theta=\mu_kmg+\mu_kFcos\theta[/tex]
[tex]Fsin\theta-\mu_kFcos\theta=\mu_kmg[/tex]
[tex]F(sin\theta-\mu_kcos\theta)=\mu_kmg[/tex]
[tex]F=\frac{\mu_kmg}{sin\theta-\mu_kcos\theta}[/tex]
Power,P=Fv
[tex]P=\frac{\mu_kmg}{sin\theta-\mu_kcos\theta}v[/tex]
Where [tex]g=9.8m/s^2[/tex]
B.Substitute the values
[tex]P=\frac{0.16\times 12\times 9.8}{sin53-0.16cos53}\times 1.5[/tex]
[tex]P=40.17W[/tex]
The Power the kid supplies to the lawnmower will be "40.17 W". To understand the calculation, check below.
According to the question,
Mass, m = 12 kg
Angle, θ = 53°
Speed, v = 1.5 m/s
[tex]\mu_k[/tex] = 0.16
When the net force is in x-direction,
→ Net force, [tex]F_{net}[/tex] = 0
[tex]F_{Sin \Theta}[/tex] - f = 0
[tex]F_{Sin \Theta}[/tex] = f
When the net force is in y-direction,
→ N - mg - F Cosθ = 0
or,
f = [tex]\mu_k[/tex]N = [tex]\mu_k[/tex] (mg + F Cosθ)
F Sin θ = [tex]\mu_k[/tex] (mg + F Cosθ)
F = [tex]\frac{\mu_k mg}{Sin \theta - \mu_k Cos \theta}[/tex]
We know, Power, P = Fv
then, P = [tex]\frac{\mu_k mg}{Sin \theta - \mu_k Cos \theta}[/tex] v
By substituting the values,
= [tex]\frac{0.16\times 12\times 9.8}{Sin 53^{\circ} - 0.16 Cos 53^{\circ}}[/tex] × 1.5
= 40.17 W
Thus the approach above is correct.
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