A 3kg mass is sliding across a horizontal surface at a constant speed while being pulled by rope with tension 20N held at a 53 degree angle above the horizontal. The force of friction is most nearly
A. 30N
B. 12N
C. 16N
D. 20N
E. 1N
Tx = 20*Cos53 = 12 N. = Hor. component
of tension.
Tx-Fk = M*a
12-Fk = M*0 = 0
Fk = 12 N.
To find the force of friction acting on the 3kg mass, we need to understand the forces acting on it.
1. The force of tension in the rope, which is 20N, can be divided into horizontal and vertical components.
- The horizontal component is given by T * cos θ, where T is the tension and θ is the angle above the horizontal. So, the horizontal component is: 20N * cos(53°).
- The vertical component is given by T * sin θ. So, the vertical component is: 20N * sin(53°).
2. The force of gravity acting on the mass is given by the formula m * g, where m is the mass and g is the acceleration due to gravity. So, the force of gravity is: 3kg * 9.8m/s^2 (approximated value of g).
3. Since the mass is sliding at a constant speed, the force of friction must be equal in magnitude and opposite in direction to the horizontal component of the tension.
Now, let's calculate the horizontal component of the tension and compare it with the answer choices.
Horizontal component of tension = 20N * cos(53°) = 20N * 0.6 (approximation) ≈ 12N
Based on this calculation, we can see that the force of friction must be approximately 12N. Therefore, the closest answer choice is B. 12N.