A box of mass 1kg is placed on a table and is connected to a hanging mass m by a string strung over a pulley. If the coefficient of kinetic friction between the box and the table is .3 and the box slides at a constant speed of .5m/s what is the weight of the hanging mass?
to find the weight of the hanging mass:
F_g=mu*mg
=.3*m*9.8
=2.94N
b) what is the tension in the string?
F_f=mu * FN
=.3*9.8
=2.94
T-F_f=ma
T-2.94= 1kg(a)
what is the accleration is it zero?
so would that mean
tension equals 2.94N (the same as the frictional force?
To find the weight of the hanging mass (m), we can use the formula F_g = mu * mg, where F_g is the force of gravity, mu is the coefficient of kinetic friction, and m and g are the mass of the object and the acceleration due to gravity, respectively.
In this case, mu is given as 0.3, and g is approximately 9.8 m/s^2. Plugging these values into the formula, we get:
F_g = 0.3 * m * 9.8
Now, since the box slides at a constant speed, we can assume that the net external force acting on the box is zero. This means that the force of friction (F_f) must be equal to the tension in the string (T). So, to find the tension in the string, we can use the formula F_f = mu * FN, where FN is the normal force exerted on the box.
Since the box is not accelerating vertically, the normal force is equal to the weight of the box, which is m * g. Therefore:
F_f = 0.3 * m * g
Now, if we consider the motion of the box horizontally, we can use Newton's second law: T - F_f = ma, where T is the tension in the string, F_f is the force of friction, m is the mass of the box, and a is the acceleration of the box.
In this case, the box is sliding at a constant speed, so the acceleration (a) is zero. Therefore:
T - F_f = 0
Since T = F_f, the tension in the string is equal to the force of friction, which is 2.94 N.
So, to answer your question, the weight of the hanging mass (m) is 2.94 N and the tension in the string is also 2.94 N.