Question

1. victoria slides a 1.1 kg stone across a frozen pond. the coefficient of kinetic friction between the stone and the ice is 0.12. what is the stones rate of acceleration as it slides across the pond?

( vec { a } = - 1.31 mathrm { m } / mathrm { s } ^ { 2 } )

( vec { a } = - 2.24 mathrm { m } / mathrm { s } ^ { 2 } )

( vec { a } = - 1.18 mathrm { m } / mathrm { s } ^ { 2 } )

( vec { a } = - 0.664 mathrm { m } / mathrm { s } ^ { 2 } )

Explanation:

Step1: Find the normal force

The normal force \(N\) on the stone is equal to its weight since there is no vertical acceleration. Using \(N = mg\), where \(m = 1.1\space kg\) and \(g=9.8\space m/s^{2}\).
\(N=(1.1\space kg)\times(9.8\space m/s^{2}) = 10.78\space N\)

Step2: Calculate the frictional force

The frictional force \(F_f=\mu_kN\), with \(\mu_k = 0.12\) and \(N = 10.78\space N\).
\(F_f=(0.12)\times(10.78\space N)=1.2936\space N\)

Step3: Apply Newton's second law \(F = ma\)

Here, the net force is the frictional force (\(F_{net}=F_f\)), and we want to find \(a\). Rearranging \(F = ma\) gives \(a=\frac{F_{net}}{m}\). Substituting \(F_{net}=- 1.2936\space N\) (negative because it opposes motion) and \(m = 1.1\space kg\)
\(a=\frac{-1.2936\space N}{1.1\space kg}=-1.176\approx - 1.18\space m/s^{2}\)

Answer:

\(\vec{a}=-1.18\space m/s^{2}\) (the third option)