The problems of variable force doing work in high school are mainly divided into the following categories, and the following are the treatment methods.
The first category: forces with constant magnitude but changing direction at any time, such as friction resistance and air resistance (constant magnitude), are applied with the formula W=fS path (S path represents the distance traveled by the object).
In high school, the force with constant magnitude and changing direction is mainly friction or air resistance (force with constant magnitude). Our general idea to solve this kind of problems is to get a universally applicable formula through "differential element method", and then students can directly apply this formula in their later study. Next, we use the "infinitesimal method" to solve the work problem of friction or air resistance (constant force).
An object moves in a curve under a rough horizontal plane, and the curve trajectory is as shown in figure 1, moving from point O to point O? Point. Now we divide the curve into many small segments, such as AB segment and CD segment, and each segment is small enough to be considered as a straight line. The time for an object to pass through each short period is short enough. In such a short time, the change of force is very small and can be considered as a constant. Therefore, for each short segment, the work can be calculated by the formula W=FScosθ. Because the direction of the resistance F is always opposite to the direction of the object's motion, the work done by the resistance F in the whole process is W = W1+W2+W3+...+WN = FS1+FS2+...+FSN = F (S1+S2+S3+... Of course, according to mathematical knowledge, if we want to establish the above formula, F must be a constant force, otherwise the common factor of F cannot be extracted from the above formula, so we can usually use the formula W=fS to find the work done by the constant force resistance, provided that the size of F remains unchanged, as shown in the following question:
For example, if a ball is thrown vertically upward at a certain initial speed, the maximum height of the ball is h and the air resistance is constant f, then the work done by the air resistance to the ball from the dropping point to the original starting point is ().
A 0 B -fh C -2fh D -4fh
Analysis: Many students think that the displacement through which an object passes is 0, so according to the formula W=FScosθ, the work done by resistance is 0, so a is chosen. In fact, because the direction of air resistance changes from the throwing point to the original starting point, the direction of air resistance is vertical downward when rising and vertical upward when falling, so it is a variable force in the whole process. Just because the displacement of an object is zero, we can't simply think that the work done by resistance F is zero. According to the law of the first kind of force doing work, we can know that the distance through which an object passes is 2h, so the air resistance does work is -2fh, that is, c is correct.
The second category: changing force into constant force.
For example, as shown in Figure 2, on a platform with a height of h=2m, a person exerts a horizontal constant force of F=80N on one end of a light rope, and the rope pulls an object on the ground from point A to point B through a smooth crown block, so as to ask for the work done by people on the object.
Analysis: Because the direction of the rope is always changing (the pulling force of the rope on the right side of the pulley), although the pulling force is constant in the process of pulling the object, we can't directly solve it with the formula W=FScosθ, so it is difficult for us to find out the work that the pulling force does on the object, but we can consider it from the left side of the pulley, because according to the law of conservation of energy, the work that people do on the object is also equal to the product of the horizontal pulling force and the distance that the rope moves in the horizontal direction. In this way, we change the problem of variable force doing work into the problem of constant force doing work. Because the distance that the pulling point moves is s = sac-SBC = h/sin30-h/sin53, we can get the work that people do to objects:
W=F(SAC-SBC)=F(h/sin30 - h/sin53)
Substitute the data to get W= 120J.
Induction: For a light rope, one end is subjected to a constant force, the middle passes through the pulley, and the other end pulls the object forward, and the pulling force is constant and the direction changes. As for the work done by this variable force, we can usually deal with it by converting it into constant force.
The third category: using kinetic energy theorem to solve.
We usually use the kinetic energy theorem to solve the work done by a force whose magnitude and direction are constantly changing, because this theorem does not need to consider the motion process of an object, but only emphasizes the initial state and final state of the object. For example:
For example, an object with mass m slides down from the top of a rough curved surface (as shown in Figure 3), the height of the curved surface is H, and the speed of the object sliding to the bottom is V, excluding air resistance. How much work does an object do to overcome resistance when it slides from the top to the bottom of a curved surface?
Analysis: Because the magnitude and direction of resistance are always changing during falling, we can't directly solve it according to the calculation formula of work, but it is often more convenient to use kinetic energy theorem. Because in the process of falling, only gravity and resistance are doing work, the gravity work is WG=mgh, and the resistance work is Wf. According to the kinetic energy theorem, WG-Wf=
1/2mv2, so Wf=WG- 1/2mv2=mgh-
1/2mv2 .
Induction: We usually use the kinetic energy theorem to deal with the work done by forces with varying magnitude and direction.
(Author: Zhejiang Daishan Middle School)
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