The force diagram above shows a box accelerating to the right on a horizontal surface of negligible friction. The tension TT is exerted at an angle of 30°30° above the horizontal. If μμ is the coefficient of kinetic friction between the box and the surface, which of the following is a correct mathematical equation derived by applying Newton’s second law to the box? a=(Tcos(θ)−μMg)/M a=(Tsin(θ)−μMg)/M a=Tcos(θ)−μ(Mg+Tsin(θ))/M a=Tcos(θ)−μ(Mg−Tsin(θ))/M
The force diagram above shows a box accelerating to the right on a horizontal surface of negligible friction. The tension TT is exerted at an angle of 30°30° above the horizontal. If μμ is the coefficient of kinetic friction between the box and the surface, which of the following is a correct mathematical equation derived by applying Newton’s second law to the box? a=(Tcos(θ)−μMg)/M a=(Tsin(θ)−μMg)/M a=Tcos(θ)−μ(Mg+Tsin(θ))/M a=Tcos(θ)−μ(Mg−Tsin(θ))/M
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The force diagram above shows a box accelerating to the right on a horizontal surface of negligible friction. The tension TT is exerted at an angle of 30°30° above the horizontal. If μμ is the coefficient of kinetic friction between the box and the surface, which of the following is a correct mathematical equation derived by applying Newton’s second law to the box?
- a=(Tcos(θ)−μMg)/M
- a=(Tsin(θ)−μMg)/M
-
a=Tcos(θ)−μ(Mg+Tsin(θ))/M
- a=Tcos(θ)−μ(Mg−Tsin(θ))/M
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