I posted this before, but my description was vague so I have a link to the picture. Please help. I don't know what to do with the static and kinetic forces.
pic.twitter.com/Un8Nra4c
Thanks
pic.twitter.com/Un8Nra4c
Thanks
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Here are the parts you were asking about (how to find normal force and acceleration).
a)
Weight = mg = 2.0 x 9.8 = 19.6N
Component of weight perpendicular to slope = Wcosθ = 19.6cos(30⁰) = 17N
Component of weight parallel to slope = Wsinθ = 19.6sin(30⁰) = 9.8N
Since there is no acceleration perpendicular to the slope, the magnitude of Fn = the magnitude of the component of weight perpendicular to slope.
Fn = 17.0N
b)
The acceleration (parallel to the slope) is produced by the resultant of Fapplied and friction (Ff). As the object is moving, we use the coefficient of kinetic friction, μk.
Frictional force = μk x Fn = 0.14 x 17.0 = 2.38N
Resultant force (parallel to slope) = 5.2 – 2.38 = 2.82N
Since F=ma, a = F/m = 2.82/2 = 1.4m/s² (to 2 significant figures)
You might find my video in the link useful – at about 18mins 45s it covers friction on a slope.
a)
Weight = mg = 2.0 x 9.8 = 19.6N
Component of weight perpendicular to slope = Wcosθ = 19.6cos(30⁰) = 17N
Component of weight parallel to slope = Wsinθ = 19.6sin(30⁰) = 9.8N
Since there is no acceleration perpendicular to the slope, the magnitude of Fn = the magnitude of the component of weight perpendicular to slope.
Fn = 17.0N
b)
The acceleration (parallel to the slope) is produced by the resultant of Fapplied and friction (Ff). As the object is moving, we use the coefficient of kinetic friction, μk.
Frictional force = μk x Fn = 0.14 x 17.0 = 2.38N
Resultant force (parallel to slope) = 5.2 – 2.38 = 2.82N
Since F=ma, a = F/m = 2.82/2 = 1.4m/s² (to 2 significant figures)
You might find my video in the link useful – at about 18mins 45s it covers friction on a slope.