What is the angle of the slope in Q2?

[lemon]

Q1. A cyclist can develop power of 400W and a top speed on a level road of 7.5m/s. The drag force against which the cyclist is working is?
Q2. A constant force drags a block of mass 2kg up a smooth slope at a constant speed of 1m/s, exerting power or 10W. The angle of the slope is?



2. P=F x V
mgsin(theta)
F=ma



3Q1. 400W/7.5m/s = 53.3N (3s.f.)

This is just a simple solution stating that the drag force is equal to the forward thrust force, right? As not mention of Frictional Force has been mentioned.

3Q2. If there is a constant force there is no acceleration.
Force acting up the slope - P/V = 10/1 = 10N
Force acting downwards - mgsin(theta). Now I'm stuck. I think I need to calculate F-F₁=mgsin(theta). But I have two missing numbers for this equation. If I try to find acceleration using F=ma I get 10N/2kg = 5ms^-2. But if this is a constant force this is no acceleration.
Confused. Please help?

 

[Chewy0087]

First question looks fine, it's a level surface, so the "Drag Force" is friction, but you've answered it correctly.

For the second, remember the equation should be more clearly stated as

P = F_{net}v

and F(net) is all of the forces at once, you don't have to resolve from there, as you rightly said, the force is equal to mgsin(theta) so you can solve for theta.

 

[lemon]

F_gravity=mgsin(theta)
P=(F-mgsin(theta))v
rearranging - P/F-F= -mgsin(theta)
but this gives 10W/1m/s-10N = 0
The calculation cannot continue from here

 

[lemon]

Do you just mean that the F_netis equal to mgsin(theta)
So P/mg=sin(theta)?

 

[lemon]

This would give 30

 

[lemon]

anyone! Is this correct.
Total force up slope = Total force down slope. There is no resultant force along the slope.
so, F = mgsin(theta)
Theta = 30

 

[Chewy0087]

Looks right to me.

 

[lemon]

ok. thanks Chewy